Vector Algebra — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 10, Vector Algebra (ভেক্টৰ বীজগণিত). It covers Exercise 10.1, Exercise 10.2, Exercise 10.3, Exercise 10.4 and the Miscellaneous Exercise on Chapter 10, with each result worked out in full and figures drawn wherever a question depends on one, so you can follow the reasoning and prepare confidently for your examination.
Summary
A scalar has only magnitude (length, mass, time), while a vector has both magnitude and direction (displacement, velocity, force). The position vector of a point $P(x, y, z)$ is $\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$, with magnitude $|\vec{OP}| = \sqrt{x^2 + y^2 + z^2}$. The direction cosines $l, m, n$ (cosines of the angles $\alpha, \beta, \gamma$ the vector makes with the axes) satisfy $l^2 + m^2 + n^2 = 1$, and for a vector with direction ratios $a, b, c$ and magnitude $r$ we have $l = \frac{a}{r}$, $m = \frac{b}{r}$, $n = \frac{c}{r}$. The unit vector along $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
Vectors add by the triangle and parallelogram laws; in component form $\vec{a} \pm \vec{b}$ adds or subtracts corresponding components, and $\lambda\vec{a}$ scales each component by $\lambda$. The point $R$ dividing the join of $P$ and $Q$ (position vectors $\vec{a}, \vec{b}$) in the ratio $m : n$ has position vector $\frac{m\vec{b} + n\vec{a}}{m + n}$ (internally) and $\frac{m\vec{b} – n\vec{a}}{m – n}$ (externally). The scalar (dot) product is $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1 b_1 + a_2 b_2 + a_3 b_3$, giving $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ and the projection of $\vec{a}$ on $\vec{b}$ as $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
The vector (cross) product is $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}$, a vector perpendicular to both $\vec{a}$ and $\vec{b}$; in components it is the determinant with rows $\hat{i}, \hat{j}, \hat{k}$; $a_1, a_2, a_3$; $b_1, b_2, b_3$. Its magnitude $|\vec{a} \times \vec{b}|$ equals the area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$, so the area of a triangle on those sides is $\frac{1}{2}|\vec{a} \times \vec{b}|$. Two vectors are perpendicular when $\vec{a} \cdot \vec{b} = 0$ and parallel (collinear) when $\vec{a} \times \vec{b} = \vec{0}$.
Summary: ASSEB Class 12 Mathematics Chapter 10 Vector Algebra covers scalars and vectors, position vectors and direction cosines, addition and scalar multiplication, the section formula, and the scalar (dot) and vector (cross) products with their use in finding angles, projections, and areas, with complete worked answers to Exercise 10.1, 10.2, 10.3, 10.4 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 10.1
1. Represent graphically a displacement of 40 km, 30° east of north.
Answer: Take the starting point $O$. “30° east of north” means the direction obtained by turning $30°$ from north towards the east. The displacement is drawn as a vector $\vec{OP}$ from $O$ in that direction, whose length represents $40$ km on a chosen scale.
2. Classify the following measures as scalars and vectors.
(i) 10 kg (ii) 2 meters north-west (iii) 40° (iv) 40 watt (v) $10^{-19}$ coulomb (vi) 20 m/s²
Answer: A quantity is a vector only if a direction is attached to its magnitude.
- (i) 10 kg — scalar (mass; only magnitude).
- (ii) 2 meters north-west — vector (a distance with a stated direction).
- (iii) 40° — scalar (a plain angle).
- (iv) 40 watt — scalar (power).
- (v) $10^{-19}$ coulomb — scalar (electric charge).
- (vi) 20 m/s² — scalar (only the magnitude of acceleration is given, with no direction).
3. Classify the following as scalar and vector quantities.
(i) time period (ii) distance (iii) force (iv) velocity (v) work done
Answer:
- (i) time period — scalar.
- (ii) distance — scalar.
- (iii) force — vector.
- (iv) velocity — vector.
- (v) work done — scalar.
4. In Fig 10.6 (a square), identify the following vectors.
(i) Coinitial (ii) Equal (iii) Collinear but not equal
Answer: In the square, $\vec{a}$ points right along the top, $\vec{b}$ points down the right side, $\vec{c}$ points left along the bottom and $\vec{d}$ points down the left side.
- (i) Coinitial vectors: $\vec{a}$ and $\vec{d}$ — both start from the top-left corner.
- (ii) Equal vectors: $\vec{b}$ and $\vec{d}$ — both point vertically downward and have the same length (a side of the square).
- (iii) Collinear but not equal: $\vec{a}$ and $\vec{c}$ — both are horizontal (hence collinear) and equal in length, but point in opposite directions, so they are not equal.
5. Answer the following as true or false.
(i) $\vec{a}$ and $-\vec{a}$ are collinear. (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal.
Answer:
- (i) True. $-\vec{a}$ is parallel to $\vec{a}$ (opposite direction, same supporting line), so they are collinear.
- (ii) False. Collinear vectors are only parallel; their magnitudes may differ, e.g. $\vec{a}$ and $3\vec{a}$.
- (iii) False. Equal magnitude does not force the same (or parallel) direction; the vectors may point along different lines.
- (iv) False. They may point in opposite directions, e.g. $\vec{a}$ and $-\vec{a}$ are collinear with equal magnitude but are not equal.
Exercise 10.2
1. Compute the magnitude of the following vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$; $\vec{b} = 2\hat{i} – 7\hat{j} – 3\hat{k}$; $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} – \frac{1}{\sqrt{3}}\hat{k}$.
Answer: The magnitude of $x\hat{i} + y\hat{j} + z\hat{k}$ is $\sqrt{x^2 + y^2 + z^2}$.
$$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$
$$|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$$
$$|\vec{c}| = \sqrt{\tfrac{1}{3} + \tfrac{1}{3} + \tfrac{1}{3}} = \sqrt{1} = 1$$
2. Write two different vectors having same magnitude.
Answer: Take $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} + \hat{k}$. Then $|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$ and $|\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14}$. The two vectors are different (their components differ) yet have the same magnitude $\sqrt{14}$.
3. Write two different vectors having same direction.
Answer: Take $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k}$. Since $\vec{b} = 2\vec{a}$, both have the same unit vector $\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$, hence the same direction, but they are different vectors (different magnitudes).
4. Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.
Answer: Two vectors are equal iff their corresponding components are equal. Comparing, $x = 2$ and $y = 3$.
5. Find the scalar and vector components of the vector with initial point $(2, 1)$ and terminal point $(-5, 7)$.
Answer: The vector is (terminal $-$ initial):
$$\vec{PQ} = (-5 – 2)\hat{i} + (7 – 1)\hat{j} = -7\hat{i} + 6\hat{j}$$
The scalar components are $-7$ and $6$; the vector components are $-7\hat{i}$ and $6\hat{j}$.
6. Find the sum of the vectors $\vec{a} = \hat{i} – 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} – 6\hat{j} – 7\hat{k}$.
Answer: Add corresponding components:
$$\vec{a} + \vec{b} + \vec{c} = (1 – 2 + 1)\hat{i} + (-2 + 4 – 6)\hat{j} + (1 + 5 – 7)\hat{k} = -4\hat{j} – \hat{k}$$
7. Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.
Answer: $|\vec{a}| = \sqrt{1 + 1 + 4} = \sqrt{6}$. Hence
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k}) = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$$
8. Find the unit vector in the direction of vector $\vec{PQ}$, where $P$ and $Q$ are the points $(1, 2, 3)$ and $(4, 5, 6)$, respectively.
Answer: $\vec{PQ} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$, so $|\vec{PQ}| = \sqrt{9 + 9 + 9} = 3\sqrt{3}$.
$$\widehat{PQ} = \frac{1}{3\sqrt{3}}(3\hat{i} + 3\hat{j} + 3\hat{k}) = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$$
9. For given vectors, $\vec{a} = 2\hat{i} – \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} – \hat{k}$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.
Answer: $\vec{a} + \vec{b} = (2-1)\hat{i} + (-1+1)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{k}$, with $|\vec{a} + \vec{b}| = \sqrt{1 + 0 + 1} = \sqrt{2}$.
$$\widehat{(\vec{a} + \vec{b})} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}) = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$$
10. Find a vector in the direction of vector $5\hat{i} – \hat{j} + 2\hat{k}$ which has magnitude 8 units.
Answer: The magnitude of $5\hat{i} – \hat{j} + 2\hat{k}$ is $\sqrt{25 + 1 + 4} = \sqrt{30}$. A vector of magnitude $8$ in this direction is $8$ times the unit vector:
$$\frac{8}{\sqrt{30}}(5\hat{i} – \hat{j} + 2\hat{k}) = \frac{40}{\sqrt{30}}\hat{i} – \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$$
11. Show that the vectors $2\hat{i} – 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} – 8\hat{k}$ are collinear.
Answer: Let $\vec{a} = 2\hat{i} – 3\hat{j} + 4\hat{k}$ and $\vec{b} = -4\hat{i} + 6\hat{j} – 8\hat{k}$. Then $\vec{b} = -2(2\hat{i} – 3\hat{j} + 4\hat{k}) = -2\vec{a}$. Since $\vec{b} = \lambda\vec{a}$ with the scalar $\lambda = -2$, the two vectors are collinear (parallel).
12. Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.
Answer: Magnitude $= \sqrt{1 + 4 + 9} = \sqrt{14}$. The direction cosines are the components divided by the magnitude:
$$l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}$$
13. Find the direction cosines of the vector joining the points $A(1, 2, -3)$ and $B(-1, -2, 1)$, directed from $A$ to $B$.
Answer: $\vec{AB} = (-1-1)\hat{i} + (-2-2)\hat{j} + (1+3)\hat{k} = -2\hat{i} – 4\hat{j} + 4\hat{k}$, with $|\vec{AB}| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$$l = \frac{-2}{6} = -\frac{1}{3}, \quad m = \frac{-4}{6} = -\frac{2}{3}, \quad n = \frac{4}{6} = \frac{2}{3}$$
14. Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ.
Answer: Magnitude $= \sqrt{1 + 1 + 1} = \sqrt{3}$, so the direction cosines are
$$\cos\alpha = \frac{1}{\sqrt{3}}, \quad \cos\beta = \frac{1}{\sqrt{3}}, \quad \cos\gamma = \frac{1}{\sqrt{3}}$$
Since $\cos\alpha = \cos\beta = \cos\gamma$, the angles are equal, $\alpha = \beta = \gamma = \cos^{-1}\frac{1}{\sqrt{3}}$. Hence the vector is equally inclined to the three axes.
15. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\hat{i} + 2\hat{j} – \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio $2 : 1$ (i) internally (ii) externally.
Answer: Let $\vec{p} = \hat{i} + 2\hat{j} – \hat{k}$ and $\vec{q} = -\hat{i} + \hat{j} + \hat{k}$, with $m = 2$, $n = 1$.
(i) Internal division: $\vec{r} = \dfrac{m\vec{q} + n\vec{p}}{m + n}$.
$$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} – \hat{k})}{3} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$$
(ii) External division: $\vec{r} = \dfrac{m\vec{q} – n\vec{p}}{m – n}$.
$$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) – (\hat{i} + 2\hat{j} – \hat{k})}{2 – 1} = -3\hat{i} + 3\hat{k}$$
16. Find the position vector of the mid point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.
Answer: The mid point has position vector $\frac{\vec{p} + \vec{q}}{2}$:
$$\frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} – 2\hat{k})}{2} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$$
17. Show that the points $A$, $B$ and $C$ with position vectors $\vec{a} = 3\hat{i} – 4\hat{j} – 4\hat{k}$, $\vec{b} = 2\hat{i} – \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} – 3\hat{j} – 5\hat{k}$, respectively form the vertices of a right angled triangle.
Answer: Form the side vectors:
$$\vec{AB} = \vec{b} – \vec{a} = -\hat{i} + 3\hat{j} + 5\hat{k}, \quad |\vec{AB}|^2 = 1 + 9 + 25 = 35$$
$$\vec{BC} = \vec{c} – \vec{b} = -\hat{i} – 2\hat{j} – 6\hat{k}, \quad |\vec{BC}|^2 = 1 + 4 + 36 = 41$$
$$\vec{CA} = \vec{a} – \vec{c} = 2\hat{i} – \hat{j} + \hat{k}, \quad |\vec{CA}|^2 = 4 + 1 + 1 = 6$$
Since $|\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41 = |\vec{BC}|^2$, the triangle satisfies the Pythagoras relation with the right angle at $A$. Hence $A, B, C$ form a right angled triangle.
18. In triangle ABC (Fig 10.18), which of the following is not true:
(A) $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$ (B) $\vec{AB} + \vec{BC} – \vec{AC} = \vec{0}$ (C) $\vec{AB} + \vec{BC} – \vec{CA} = \vec{0}$ (D) $\vec{AB} – \vec{CB} + \vec{CA} = \vec{0}$
Answer: (C). By the triangle law, $\vec{AB} + \vec{BC} = \vec{AC}$, so $\vec{AB} + \vec{BC} + \vec{CA} = \vec{AC} + \vec{CA} = \vec{0}$; option (A) is true. Option (B): $\vec{AB} + \vec{BC} – \vec{AC} = \vec{AC} – \vec{AC} = \vec{0}$, true. Option (D): since $-\vec{CB} = \vec{BC}$, $\vec{AB} – \vec{CB} + \vec{CA} = \vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$, true. But (C) gives $\vec{AB} + \vec{BC} – \vec{CA} = \vec{AC} – \vec{CA} = \vec{AC} + \vec{AC} = 2\vec{AC} \neq \vec{0}$. Hence (C) is not true.
19. If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:
(A) $\vec{b} = \lambda\vec{a}$, for some scalar $\lambda$ (B) $\vec{a} = \pm\vec{b}$ (C) the respective components of $\vec{a}$ and $\vec{b}$ are not proportional (D) both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes.
Answer: (B), (C) and (D) are incorrect. For collinear vectors there always exists a scalar $\lambda$ with $\vec{b} = \lambda\vec{a}$, so (A) is correct. Statement (B) is false because $\vec{a} = \pm\vec{b}$ only when the magnitudes are equal. Statement (C) is false because collinear vectors have proportional components. Statement (D) is false because collinear vectors may point in opposite directions and may have equal magnitudes. Thus the incorrect statements are (B), (C) and (D).
Exercise 10.3
1. Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 2, respectively having $\vec{a} \cdot \vec{b} = \sqrt{6}$.
Answer: Using $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$,
$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3} \cdot 2} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}}$$
Hence $\theta = \frac{\pi}{4}$ (i.e. $45°$).
2. Find the angle between the vectors $\hat{i} – 2\hat{j} + 3\hat{k}$ and $3\hat{i} – 2\hat{j} + \hat{k}$.
Answer: $\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$; $|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$ and $|\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14}$.
$$\cos\theta = \frac{10}{\sqrt{14}\cdot\sqrt{14}} = \frac{10}{14} = \frac{5}{7} \quad\Rightarrow\quad \theta = \cos^{-1}\frac{5}{7}$$
3. Find the projection of the vector $\hat{i} – \hat{j}$ on the vector $\hat{i} + \hat{j}$.
Answer: Projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. Here $\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 0$, so the projection is $\frac{0}{\sqrt{2}} = 0$. (The two vectors are perpendicular.)
4. Find the projection of the vector $\hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $7\hat{i} – \hat{j} + 8\hat{k}$.
Answer: $\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 – 3 + 56 = 60$ and $|\vec{b}| = \sqrt{49 + 1 + 64} = \sqrt{114}$.
$$\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{60}{\sqrt{114}}$$
5. Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k})$, $\frac{1}{7}(3\hat{i} – 6\hat{j} + 2\hat{k})$, $\frac{1}{7}(6\hat{i} + 2\hat{j} – 3\hat{k})$. Also, show that they are mutually perpendicular to each other.
Answer: Call them $\vec{a}, \vec{b}, \vec{c}$. Their magnitudes are
$$|\vec{a}| = \frac{1}{7}\sqrt{4 + 9 + 36} = \frac{\sqrt{49}}{7} = 1, \quad |\vec{b}| = \frac{1}{7}\sqrt{9 + 36 + 4} = 1, \quad |\vec{c}| = \frac{1}{7}\sqrt{36 + 4 + 9} = 1$$
So each is a unit vector. For perpendicularity, check the dot products:
$$\vec{a} \cdot \vec{b} = \frac{1}{49}(6 – 18 + 12) = 0, \quad \vec{b} \cdot \vec{c} = \frac{1}{49}(18 – 12 – 6) = 0, \quad \vec{c} \cdot \vec{a} = \frac{1}{49}(12 + 6 – 18) = 0$$
All three dot products vanish, so the vectors are mutually perpendicular.
6. Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.
Answer: $(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = |\vec{a}|^2 – |\vec{b}|^2 = 8$. Put $|\vec{a}| = 8|\vec{b}|$:
$$64|\vec{b}|^2 – |\vec{b}|^2 = 8 \;\Rightarrow\; 63|\vec{b}|^2 = 8 \;\Rightarrow\; |\vec{b}|^2 = \frac{8}{63}$$
$$|\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}} = \frac{2\sqrt{14}}{21}, \quad |\vec{a}| = 8|\vec{b}| = \frac{16\sqrt{14}}{21}$$
7. Evaluate the product $(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$.
Answer: Expand using distributivity and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$:
$$(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}) = 6(\vec{a}\cdot\vec{a}) + 21(\vec{a}\cdot\vec{b}) – 10(\vec{b}\cdot\vec{a}) – 35(\vec{b}\cdot\vec{b})$$
$$= 6|\vec{a}|^2 + 11\,\vec{a}\cdot\vec{b} – 35|\vec{b}|^2$$
8. Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude and such that the angle between them is 60° and their scalar product is $\frac{1}{2}$.
Answer: Let $|\vec{a}| = |\vec{b}| = t$. Then $\vec{a} \cdot \vec{b} = t \cdot t \cos 60° = t^2 \cdot \frac{1}{2} = \frac{1}{2}$, so $t^2 = 1$ and $t = 1$. Hence $|\vec{a}| = |\vec{b}| = 1$.
9. Find $|\vec{x}|$, if for a unit vector $\vec{a}$, $(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$.
Answer: $(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a}) = |\vec{x}|^2 – |\vec{a}|^2 = 12$. Since $\vec{a}$ is a unit vector, $|\vec{a}|^2 = 1$, so $|\vec{x}|^2 = 13$ and $|\vec{x}| = \sqrt{13}$.
10. If $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j}$ are such that $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda$.
Answer: $\vec{a} + \lambda\vec{b} = (2 – \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}$. Perpendicular to $\vec{c}$ means $(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$:
$$3(2 – \lambda) + 1(2 + 2\lambda) + 0(3 + \lambda) = 6 – 3\lambda + 2 + 2\lambda = 8 – \lambda = 0 \;\Rightarrow\; \lambda = 8$$
11. Show that $|\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b} – |\vec{b}|\vec{a}$, for any two nonzero vectors $\vec{a}$ and $\vec{b}$.
Answer: Take the dot product and use $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$:
$$(|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} – |\vec{b}|\vec{a}) = |\vec{a}|^2(\vec{b}\cdot\vec{b}) – |\vec{b}|^2(\vec{a}\cdot\vec{a}) = |\vec{a}|^2|\vec{b}|^2 – |\vec{b}|^2|\vec{a}|^2 = 0$$
(The cross terms $\pm|\vec{a}||\vec{b}|\,\vec{a}\cdot\vec{b}$ cancel.) Since the dot product is zero, the two vectors are perpendicular.
12. If $\vec{a} \cdot \vec{a} = 0$ and $\vec{a} \cdot \vec{b} = 0$, then what can be concluded about the vector $\vec{b}$?
Answer: $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 0$ forces $|\vec{a}| = 0$, so $\vec{a} = \vec{0}$. But $\vec{0} \cdot \vec{b} = 0$ holds for every vector $\vec{b}$. Therefore nothing special can be concluded about $\vec{b}$; it may be any vector.
13. If $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.
Answer: Square the given relation: $|\vec{a} + \vec{b} + \vec{c}|^2 = 0$, i.e.
$$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$$
Each magnitude is $1$, so $3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$, giving $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2}$.
14. If either vector $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \cdot \vec{b} = 0$. But the converse need not be true. Justify your answer with an example.
Answer: The converse would claim “$\vec{a} \cdot \vec{b} = 0 \Rightarrow \vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$”, which is false: two nonzero perpendicular vectors also have zero dot product. Take $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} – \hat{j}$. Neither is the zero vector ($|\vec{a}| = |\vec{b}| = \sqrt{2}$), yet $\vec{a} \cdot \vec{b} = 1 – 1 = 0$. So $\vec{a} \cdot \vec{b} = 0$ does not require either vector to be zero.
15. If the vertices $A$, $B$, $C$ of a triangle ABC are $(1, 2, 3)$, $(-1, 0, 0)$, $(0, 1, 2)$, respectively, then find $\angle ABC$. [$\angle ABC$ is the angle between the vectors $\vec{BA}$ and $\vec{BC}$.]
Answer: $\vec{BA} = A – B = 2\hat{i} + 2\hat{j} + 3\hat{k}$, so $|\vec{BA}| = \sqrt{4 + 4 + 9} = \sqrt{17}$. $\vec{BC} = C – B = \hat{i} + \hat{j} + 2\hat{k}$, so $|\vec{BC}| = \sqrt{1 + 1 + 4} = \sqrt{6}$. Also $\vec{BA} \cdot \vec{BC} = 2 + 2 + 6 = 10$.
$$\cos(\angle ABC) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}||\vec{BC}|} = \frac{10}{\sqrt{17}\cdot\sqrt{6}} = \frac{10}{\sqrt{102}} \;\Rightarrow\; \angle ABC = \cos^{-1}\frac{10}{\sqrt{102}}$$
16. Show that the points $A(1, 2, 7)$, $B(2, 6, 3)$ and $C(3, 10, -1)$ are collinear.
Answer: $\vec{AB} = \hat{i} + 4\hat{j} – 4\hat{k}$, so $|\vec{AB}| = \sqrt{1 + 16 + 16} = \sqrt{33}$. $\vec{BC} = \hat{i} + 4\hat{j} – 4\hat{k}$, so $|\vec{BC}| = \sqrt{33}$. $\vec{AC} = 2\hat{i} + 8\hat{j} – 8\hat{k}$, so $|\vec{AC}| = \sqrt{4 + 64 + 64} = \sqrt{132} = 2\sqrt{33}$.
Since $|\vec{AB}| + |\vec{BC}| = \sqrt{33} + \sqrt{33} = 2\sqrt{33} = |\vec{AC}|$ (and indeed $\vec{AB} = \vec{BC}$), the points $A$, $B$, $C$ are collinear.
17. Show that the vectors $2\hat{i} – \hat{j} + \hat{k}$, $\hat{i} – 3\hat{j} – 5\hat{k}$ and $3\hat{i} – 4\hat{j} – 4\hat{k}$ form the vertices of a right angled triangle.
Answer: Let $A(2, -1, 1)$, $B(1, -3, -5)$, $C(3, -4, -4)$ be the points with these position vectors.
$$\vec{AB} = -\hat{i} – 2\hat{j} – 6\hat{k},\; |\vec{AB}|^2 = 41; \quad \vec{BC} = 2\hat{i} – \hat{j} + \hat{k},\; |\vec{BC}|^2 = 6; \quad \vec{CA} = -\hat{i} + 3\hat{j} + 5\hat{k},\; |\vec{CA}|^2 = 35$$
Since $|\vec{BC}|^2 + |\vec{CA}|^2 = 6 + 35 = 41 = |\vec{AB}|^2$, the Pythagoras relation holds with the right angle at $C$. Hence the three points form a right angled triangle.
18. If $\vec{a}$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is unit vector if
(A) $\lambda = 1$ (B) $\lambda = -1$ (C) $a = |\lambda|$ (D) $a = 1/|\lambda|$
Answer: (D). $\lambda\vec{a}$ is a unit vector iff $|\lambda\vec{a}| = |\lambda|\,|\vec{a}| = |\lambda|\,a = 1$, which gives $a = \frac{1}{|\lambda|}$.
Exercise 10.4
1. Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = \hat{i} – 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k}$.
Answer:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} = \hat{i}(-14 + 14) – \hat{j}(2 – 21) + \hat{k}(-2 + 21) = 19\hat{j} + 19\hat{k}$$
$$|\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{722} = 19\sqrt{2}$$
2. Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} – \vec{b}$, where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} – 2\hat{k}$.
Answer: $\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j}$ and $\vec{a} – \vec{b} = 2\hat{i} + 4\hat{k}$. A vector perpendicular to both is their cross product:
$$(\vec{a} + \vec{b}) \times (\vec{a} – \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = 16\hat{i} – 16\hat{j} – 8\hat{k}$$
Its magnitude is $\sqrt{256 + 256 + 64} = \sqrt{576} = 24$. Hence the required unit vector is
$$\frac{1}{24}(16\hat{i} – 16\hat{j} – 8\hat{k}) = \frac{2}{3}\hat{i} – \frac{2}{3}\hat{j} – \frac{1}{3}\hat{k}$$
(The opposite vector $-\frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$ is also acceptable.)
3. If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\vec{a}$.
Answer: The direction cosines are $l = \cos\frac{\pi}{3} = \frac{1}{2}$, $m = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, $n = \cos\theta$. Using $l^2 + m^2 + n^2 = 1$:
$$\frac{1}{4} + \frac{1}{2} + \cos^2\theta = 1 \;\Rightarrow\; \cos^2\theta = \frac{1}{4} \;\Rightarrow\; \cos\theta = \frac{1}{2}$$
(taking the positive value since $\theta$ is acute), so $\theta = \frac{\pi}{3}$. The components of $\vec{a}$ are
$$\vec{a} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k}$$
4. Show that $(\vec{a} – \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$.
Answer: Expand the left side using distributivity of the cross product, and $\vec{a} \times \vec{a} = \vec{b} \times \vec{b} = \vec{0}$, $\vec{b} \times \vec{a} = -\,\vec{a} \times \vec{b}$:
$$(\vec{a} – \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a}\times\vec{a} + \vec{a}\times\vec{b} – \vec{b}\times\vec{a} – \vec{b}\times\vec{b} = \vec{0} + \vec{a}\times\vec{b} + \vec{a}\times\vec{b} – \vec{0} = 2(\vec{a}\times\vec{b})$$
5. Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.
Answer: A zero cross product means the vectors are parallel, so their components are proportional:
$$\frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu}$$
From $\frac{2}{1} = \frac{6}{\lambda}$ we get $\lambda = 3$, and from $\frac{2}{1} = \frac{27}{\mu}$ we get $\mu = \frac{27}{2}$.
6. Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?
Answer: $\vec{a} \cdot \vec{b} = 0$ means $\vec{a}$ and $\vec{b}$ are perpendicular or at least one of them is $\vec{0}$; $\vec{a} \times \vec{b} = \vec{0}$ means they are parallel or at least one is $\vec{0}$. Two nonzero vectors cannot be both perpendicular and parallel. Therefore at least one of $\vec{a}$ and $\vec{b}$ must be the zero vector.
7. Let the vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ be given as $a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then show that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
Answer: Here $\vec{b} + \vec{c} = (b_1 + c_1)\hat{i} + (b_2 + c_2)\hat{j} + (b_3 + c_3)\hat{k}$. Compute the left side:
$$\vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 + c_1 & b_2 + c_2 & b_3 + c_3 \end{vmatrix}$$
The $\hat{i}$-component is $a_2(b_3 + c_3) – a_3(b_2 + c_2) = (a_2 b_3 – a_3 b_2) + (a_2 c_3 – a_3 c_2)$, which is exactly the $\hat{i}$-component of $\vec{a}\times\vec{b}$ plus that of $\vec{a}\times\vec{c}$. The $\hat{j}$- and $\hat{k}$-components split in the same way, since each entry of the third row is a sum. Therefore
$$\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$$
8. If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is the converse true? Justify your answer with an example.
Answer: The converse “$\vec{a} \times \vec{b} = \vec{0} \Rightarrow \vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$” is not true, because two nonzero parallel vectors also give a zero cross product. Take $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} + 6\hat{j} + 2\hat{k} = 2\vec{a}$. Neither is zero, yet $\vec{a} \times \vec{b} = \vec{a} \times 2\vec{a} = 2(\vec{a}\times\vec{a}) = \vec{0}$. So $\vec{a}\times\vec{b} = \vec{0}$ does not require either vector to be zero.
9. Find the area of the triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$ and $C(1, 5, 5)$.
Answer: $\vec{AB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{AC} = 4\hat{j} + 3\hat{k}$.
$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = -6\hat{i} – 3\hat{j} + 4\hat{k}$$
$$\text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{36 + 9 + 16} = \frac{1}{2}\sqrt{61} \text{ square units}$$
10. Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} – \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} – 7\hat{j} + \hat{k}$.
Answer:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = 20\hat{i} + 5\hat{j} – 5\hat{k}$$
$$\text{Area} = |\vec{a} \times \vec{b}| = \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2} \text{ square units}$$
11. Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is
(A) $\pi/6$ (B) $\pi/4$ (C) $\pi/3$ (D) $\pi/2$
Answer: (B). $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 3 \cdot \frac{\sqrt{2}}{3} \cdot \sin\theta = \sqrt{2}\sin\theta$. For a unit vector this equals $1$, so $\sin\theta = \frac{1}{\sqrt{2}}$, giving $\theta = \frac{\pi}{4}$.
12. Area of a rectangle having vertices $A$, $B$, $C$ and $D$ with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is
(A) $\frac{1}{2}$ (B) $1$ (C) $2$ (D) $4$
Answer: (C). The adjacent sides are $\vec{AB} = 2\hat{i}$ (length $2$) and $\vec{BC} = -\hat{j}$ (length $1$), and they are perpendicular. So the area is $|\vec{AB}||\vec{BC}| = 2 \times 1 = 2$ square units.
Miscellaneous Exercise on Chapter 10
1. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of $x$-axis.
Answer: A unit vector in the XY-plane at angle $\theta$ with the $x$-axis is $\cos\theta\,\hat{i} + \sin\theta\,\hat{j}$. With $\theta = 30°$:
$$\cos 30°\,\hat{i} + \sin 30°\,\hat{j} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$$
2. Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.
Answer: $\vec{PQ} = (x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j} + (z_2 – z_1)\hat{k}$. The scalar components are $x_2 – x_1$, $y_2 – y_1$, $z_2 – z_1$, and the magnitude is
$$|\vec{PQ}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}$$
3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
Answer: Take $\hat{i}$ towards east and $\hat{j}$ towards north, and let $O$ be the start. The first walk is $\vec{OA} = -4\hat{i}$. The second, $3$ km in the direction $30°$ east of north, has east component $3\sin 30° = \frac{3}{2}$ and north component $3\cos 30° = \frac{3\sqrt{3}}{2}$, so $\vec{AB} = \frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}$. The displacement is
$$\vec{OB} = \vec{OA} + \vec{AB} = \left(-4 + \frac{3}{2}\right)\hat{i} + \frac{3\sqrt{3}}{2}\hat{j} = -\frac{5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}$$
Its magnitude is $\sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{25}{4} + \frac{27}{4}} = \sqrt{13}$ km.
4. If $\vec{a} = \vec{b} + \vec{c}$, then is it true that $|\vec{a}| = |\vec{b}| + |\vec{c}|$? Justify your answer.
Answer: It is not true in general. Equality holds only when $\vec{b}$ and $\vec{c}$ have the same direction. For example, let $\vec{b} = \hat{i}$ and $\vec{c} = \hat{j}$, so $\vec{a} = \hat{i} + \hat{j}$. Then $|\vec{a}| = \sqrt{2}$, whereas $|\vec{b}| + |\vec{c}| = 1 + 1 = 2$, and $\sqrt{2} \neq 2$. In general $|\vec{a}| \leq |\vec{b}| + |\vec{c}|$ (triangle inequality).
5. Find the value of $x$ for which $x(\hat{i} + \hat{j} + \hat{k})$ is a unit vector.
Answer: $|x(\hat{i} + \hat{j} + \hat{k})| = |x|\sqrt{3} = 1$, so $|x| = \frac{1}{\sqrt{3}}$, giving $x = \pm\frac{1}{\sqrt{3}}$.
6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a} = 2\hat{i} + 3\hat{j} – \hat{k}$ and $\vec{b} = \hat{i} – 2\hat{j} + \hat{k}$.
Answer: Resultant $\vec{a} + \vec{b} = 3\hat{i} + \hat{j}$, with $|\vec{a} + \vec{b}| = \sqrt{9 + 1} = \sqrt{10}$. A vector of magnitude $5$ in this direction is
$$5 \cdot \frac{3\hat{i} + \hat{j}}{\sqrt{10}} = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j} = \frac{3\sqrt{10}}{2}\hat{i} + \frac{\sqrt{10}}{2}\hat{j}$$
7. If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} – \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} – 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} – \vec{b} + 3\vec{c}$.
Answer: $2\vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$, $-\vec{b} = -2\hat{i} + \hat{j} – 3\hat{k}$, $3\vec{c} = 3\hat{i} – 6\hat{j} + 3\hat{k}$. Adding,
$$2\vec{a} – \vec{b} + 3\vec{c} = 3\hat{i} – 3\hat{j} + 2\hat{k}, \quad |2\vec{a} – \vec{b} + 3\vec{c}| = \sqrt{9 + 9 + 4} = \sqrt{22}$$
The required unit vector is $\frac{1}{\sqrt{22}}(3\hat{i} – 3\hat{j} + 2\hat{k})$.
8. Show that the points $A(1, -2, -8)$, $B(5, 0, -2)$ and $C(11, 3, 7)$ are collinear, and find the ratio in which $B$ divides $AC$.
Answer: $\vec{AB} = 4\hat{i} + 2\hat{j} + 6\hat{k}$, $|\vec{AB}| = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}$. $\vec{BC} = 6\hat{i} + 3\hat{j} + 9\hat{k}$, $|\vec{BC}| = \sqrt{36 + 9 + 81} = \sqrt{126} = 3\sqrt{14}$. $\vec{AC} = 10\hat{i} + 5\hat{j} + 15\hat{k}$, $|\vec{AC}| = \sqrt{100 + 25 + 225} = \sqrt{350} = 5\sqrt{14}$.
Since $|\vec{AB}| + |\vec{BC}| = 2\sqrt{14} + 3\sqrt{14} = 5\sqrt{14} = |\vec{AC}|$, the points are collinear. $B$ divides $AC$ in the ratio $|\vec{AB}| : |\vec{BC}| = 2\sqrt{14} : 3\sqrt{14} = 2 : 3$.
9. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec{a} + \vec{b})$ and $(\vec{a} – 3\vec{b})$ externally in the ratio $1 : 2$. Also, show that $P$ is the mid point of the line segment $RQ$.
Answer: With $\vec{p} = 2\vec{a} + \vec{b}$, $\vec{q} = \vec{a} – 3\vec{b}$, $m = 1$, $n = 2$, external division gives
$$\vec{r} = \frac{m\vec{q} – n\vec{p}}{m – n} = \frac{1(\vec{a} – 3\vec{b}) – 2(2\vec{a} + \vec{b})}{1 – 2} = \frac{-3\vec{a} – 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}$$
Mid point of $RQ$ has position vector $\frac{\vec{r} + \vec{q}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} – 3\vec{b})}{2} = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \vec{p}$. Hence $P$ is the mid point of $RQ$.
10. The two adjacent sides of a parallelogram are $2\hat{i} – 4\hat{j} + 5\hat{k}$ and $\hat{i} – 2\hat{j} – 3\hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.
Answer: The diagonal is the sum of the sides: $\vec{d} = 3\hat{i} – 6\hat{j} + 2\hat{k}$, with $|\vec{d}| = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$. Unit vector parallel to the diagonal:
$$\hat{d} = \frac{1}{7}(3\hat{i} – 6\hat{j} + 2\hat{k}) = \frac{3}{7}\hat{i} – \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}$$
For the area, take the cross product of the two sides:
$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = 22\hat{i} + 11\hat{j} + 0\hat{k}$$
$$\text{Area} = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5} \text{ square units}$$
11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.
Answer: If a vector is equally inclined to the three axes, its direction cosines are equal: $l = m = n$. Using $l^2 + m^2 + n^2 = 1$ gives $3l^2 = 1$, so $l = \pm\frac{1}{\sqrt{3}}$. Hence the direction cosines are $\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.
12. Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} – 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} – \hat{j} + 4\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c} \cdot \vec{d} = 15$.
Answer: A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is parallel to $\vec{a} \times \vec{b}$:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = 32\hat{i} – \hat{j} – 14\hat{k}$$
So $\vec{d} = t(32\hat{i} – \hat{j} – 14\hat{k})$ for some scalar $t$. The condition $\vec{c} \cdot \vec{d} = 15$ gives $t(2\cdot 32 + (-1)(-1) + 4\cdot(-14)) = t(64 + 1 – 56) = 9t = 15$, so $t = \frac{5}{3}$. Therefore
$$\vec{d} = \frac{5}{3}(32\hat{i} – \hat{j} – 14\hat{k}) = \frac{160}{3}\hat{i} – \frac{5}{3}\hat{j} – \frac{70}{3}\hat{k}$$
13. The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} – 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.
Answer: The sum is $(2 + \lambda)\hat{i} + 6\hat{j} – 2\hat{k}$, with magnitude $\sqrt{(2 + \lambda)^2 + 36 + 4} = \sqrt{(2 + \lambda)^2 + 40}$. The scalar product of $\hat{i} + \hat{j} + \hat{k}$ with the unit vector along the sum is
$$\frac{(2 + \lambda) + 6 – 2}{\sqrt{(2 + \lambda)^2 + 40}} = \frac{\lambda + 6}{\sqrt{(2 + \lambda)^2 + 40}} = 1$$
Squaring: $(\lambda + 6)^2 = (2 + \lambda)^2 + 40$, i.e. $\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44$, so $8\lambda = 8$ and $\lambda = 1$.
14. If $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}$, $\vec{b}$ and $\vec{c}$.
Answer: Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$ and, since they are mutually perpendicular, $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$. Put $\vec{s} = \vec{a} + \vec{b} + \vec{c}$. Then
$$|\vec{s}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 3k^2 \;\Rightarrow\; |\vec{s}| = k\sqrt{3}$$
The angle $\theta_1$ between $\vec{s}$ and $\vec{a}$ satisfies $\cos\theta_1 = \frac{\vec{s}\cdot\vec{a}}{|\vec{s}||\vec{a}|} = \frac{\vec{a}\cdot\vec{a}}{k\sqrt{3}\cdot k} = \frac{k^2}{k^2\sqrt{3}} = \frac{1}{\sqrt{3}}$. By symmetry the cosines of the angles with $\vec{b}$ and $\vec{c}$ are also $\frac{1}{\sqrt{3}}$. Since all three cosines are equal, $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}$, $\vec{b}$ and $\vec{c}$.
Note: In this printing the vector in Question 14 appears mis-set as “$\vec{c} \cdot \vec{d} = 15$” (accidentally carried over from Question 12). The intended vector, solved above, is $\vec{a} + \vec{b} + \vec{c}$.
15. Prove that $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$, if and only if $\vec{a}$, $\vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}$, $\vec{b} \neq \vec{0}$.
Answer: Expanding, $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2$. This equals $|\vec{a}|^2 + |\vec{b}|^2$ if and only if $2\,\vec{a}\cdot\vec{b} = 0$, i.e. $\vec{a}\cdot\vec{b} = 0$. As $\vec{a}$ and $\vec{b}$ are nonzero, $\vec{a}\cdot\vec{b} = 0$ holds exactly when $\vec{a} \perp \vec{b}$. Hence the stated equality is equivalent to perpendicularity.
Choose the correct answer in Exercises 16 to 19.
16. If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \geq 0$ only when
(A) $0 < \theta < \frac{\pi}{2}$ (B) $0 \leq \theta \leq \frac{\pi}{2}$ (C) $0 < \theta < \pi$ (D) $0 \leq \theta \leq \pi$
Answer: (B). $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \geq 0$ requires $\cos\theta \geq 0$, which holds for $0 \leq \theta \leq \frac{\pi}{2}$.
17. Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if
(A) $\theta = \frac{\pi}{4}$ (B) $\theta = \frac{\pi}{3}$ (C) $\theta = \frac{\pi}{2}$ (D) $\theta = \frac{2\pi}{3}$
Answer: (D). $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 1 + 2\cos\theta + 1$. For a unit vector this equals $1$, so $2 + 2\cos\theta = 1$, giving $\cos\theta = -\frac{1}{2}$ and $\theta = \frac{2\pi}{3}$.
18. The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is
(A) $0$ (B) $-1$ (C) $1$ (D) $3$
Answer: (C). $\hat{j} \times \hat{k} = \hat{i}$ so $\hat{i} \cdot \hat{i} = 1$; $\hat{i} \times \hat{k} = -\hat{j}$ so $\hat{j} \cdot (-\hat{j}) = -1$; $\hat{i} \times \hat{j} = \hat{k}$ so $\hat{k} \cdot \hat{k} = 1$. The sum is $1 – 1 + 1 = 1$.
19. If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to
(A) $0$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{2}$ (D) $\pi$
Answer: (B). $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}||\cos\theta|$ and $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$. Setting them equal gives $|\cos\theta| = \sin\theta$, i.e. $\tan\theta = 1$, so $\theta = \frac{\pi}{4}$.
Additional Questions and Answers
Multiple Choice Questions
1. The magnitude of the vector $2\hat{i} – 3\hat{j} + 6\hat{k}$ is
(A) $5$ (B) $7$ (C) $11$ (D) $\sqrt{11}$
Answer: (B) $7$. $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
2. The value of $\hat{i} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{k}$ is
(A) $0$ (B) $1$ (C) $3$ (D) $\sqrt{3}$
Answer: (C) $3$. Each dot product of a unit vector with itself is $1$, and $1 + 1 + 1 = 3$.
3. $\hat{i} \times \hat{j}$ equals
(A) $\hat{i}$ (B) $\hat{j}$ (C) $\hat{k}$ (D) $\vec{0}$
Answer: (C) $\hat{k}$. For the right-handed unit vectors, $\hat{i} \times \hat{j} = \hat{k}$.
4. If $\vec{a} \cdot \vec{b} = 0$ for two nonzero vectors $\vec{a}$ and $\vec{b}$, then the vectors are
(A) parallel (B) perpendicular (C) equal (D) collinear
Answer: (B) perpendicular. A zero dot product of nonzero vectors means $\cos\theta = 0$, i.e. $\theta = \frac{\pi}{2}$.
5. The unit vector in the direction of a nonzero vector $\vec{a}$ is
(A) $|\vec{a}|\,\vec{a}$ (B) $\frac{\vec{a}}{|\vec{a}|}$ (C) $\vec{a} \cdot \vec{a}$ (D) $|\vec{a}|$
Answer: (B) $\frac{\vec{a}}{|\vec{a}|}$. Dividing a vector by its own magnitude gives a vector of length $1$ in the same direction.
6. The area of the parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ is
(A) $\vec{a} \cdot \vec{b}$ (B) $\frac{1}{2}|\vec{a} \times \vec{b}|$ (C) $|\vec{a} \times \vec{b}|$ (D) $|\vec{a}||\vec{b}|$
Answer: (C) $|\vec{a} \times \vec{b}|$. The magnitude of the cross product is the parallelogram area; half of it is the triangle area.
7. If $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} – 2\hat{j} + 3\hat{k}$ are perpendicular, then $\lambda$ is
(A) $\frac{5}{2}$ (B) $-\frac{5}{2}$ (C) $2$ (D) $1$
Answer: (A) $\frac{5}{2}$. $\vec{a} \cdot \vec{b} = 2 – 2\lambda + 3 = 5 – 2\lambda = 0$, so $\lambda = \frac{5}{2}$.
8. The projection of $\vec{a}$ on $\vec{b}$ is
(A) $\vec{a} \cdot \vec{b}$ (B) $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ (C) $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ (D) $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Answer: (C) $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. This is the scalar projection of $\vec{a}$ along the direction of $\vec{b}$.
Fill in the Blanks
1. A vector whose magnitude is one is called a ________ vector.
Answer: unit
2. The position vector of the point $P(x, y, z)$ is ________.
Answer: $x\hat{i} + y\hat{j} + z\hat{k}$
3. If $\vec{a}$ and $\vec{b}$ are collinear, then there is a scalar $\lambda$ such that ________.
Answer: $\vec{b} = \lambda\vec{a}$
4. $\hat{j} \times \hat{k} =$ ________.
Answer: $\hat{i}$
5. If $l, m, n$ are the direction cosines of a vector, then $l^2 + m^2 + n^2 =$ ________.
Answer: $1$
True or False
1. The scalar product of two vectors is a vector.
Answer: False. The scalar (dot) product is a scalar; it is the vector (cross) product that yields a vector.
2. The cross product of two parallel vectors is the zero vector.
Answer: True. When $\vec{a}$ and $\vec{b}$ are parallel, $\sin\theta = 0$, so $\vec{a} \times \vec{b} = \vec{0}$.
3. For any two vectors, $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$.
Answer: False. The cross product is anti-commutative: $\vec{b} \times \vec{a} = -\,\vec{a} \times \vec{b}$.
4. The magnitude of a vector can be negative.
Answer: False. Magnitude is a length, so it is always non-negative.
5. Two equal vectors have the same magnitude and the same direction.
Answer: True. Equality of vectors requires both equal magnitude and equal direction, regardless of position.
Short Answer Questions
1. Find the unit vector in the direction of $\vec{a} = 3\hat{i} + 4\hat{k}$.
Answer: $|\vec{a}| = \sqrt{9 + 16} = 5$, so $\hat{a} = \frac{1}{5}(3\hat{i} + 4\hat{k}) = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{k}$.
2. Find the angle between the vectors $\hat{i} + \hat{j}$ and $\hat{i} – \hat{j}$.
Answer: Their dot product is $(1)(1) + (1)(-1) = 0$, so $\cos\theta = 0$ and $\theta = \frac{\pi}{2}$ (i.e. $90°$).
3. Find $\vec{a} \times \vec{b}$ where $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$.
Answer:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 – 0) – \hat{j}(1 – 0) + \hat{k}(1 – 0) = \hat{i} – \hat{j} + \hat{k}$$
4. If $|\vec{a}| = 3$, $|\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 6$, find the angle between $\vec{a}$ and $\vec{b}$.
Answer: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{6}{3 \times 4} = \frac{1}{2}$, so $\theta = \frac{\pi}{3}$ (i.e. $60°$).
Key Terms
| Term | Meaning |
|---|---|
| Scalar | A quantity with only magnitude, such as mass, length, time or work. |
| Vector | A quantity with both magnitude and direction, such as displacement, velocity or force. |
| Position vector | The vector $\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$ of a point $P(x, y, z)$ measured from the origin. |
| Magnitude | The length of a vector; for $x\hat{i} + y\hat{j} + z\hat{k}$ it is $\sqrt{x^2 + y^2 + z^2}$. |
| Unit vector | A vector of magnitude one; the unit vector along $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$. |
| Direction cosines | The cosines $l, m, n$ of the angles a vector makes with the axes, satisfying $l^2 + m^2 + n^2 = 1$. |
| Collinear vectors | Vectors parallel to the same line; $\vec{a}$ and $\vec{b}$ are collinear iff $\vec{b} = \lambda\vec{a}$. |
| Section formula | Position vector of $R$ dividing $PQ$ in ratio $m : n$: $\frac{m\vec{b} + n\vec{a}}{m + n}$ internally, $\frac{m\vec{b} – n\vec{a}}{m – n}$ externally. |
| Scalar (dot) product | $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1 b_1 + a_2 b_2 + a_3 b_3$; a scalar, zero when the vectors are perpendicular. |
| Projection | The projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. |
| Vector (cross) product | $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}$; a vector perpendicular to both, zero when the vectors are parallel. |
| Area (cross product) | $|\vec{a} \times \vec{b}|$ is the area of the parallelogram on $\vec{a}, \vec{b}$; $\frac{1}{2}|\vec{a} \times \vec{b}|$ is the triangle area. |