NCERT Class 12 Mathematics — Vector Algebra, Exercise 10.3. All 18 questions solved.

Exercise 10.3 introduces the scalar product, or dot product, which is the tool that lets geometry be done by arithmetic. Two definitions run the whole exercise:

$$\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta \qquad\text{and}\qquad \vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3$$

The first is geometric and the second is computational; setting them equal is how nearly every question here gets solved. Two consequences do most of the work:

• $\vec{a}\cdot\vec{b} = 0$ means the vectors are perpendicular (for non-zero vectors), since $\cos 90° = 0$.
• The projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ — note the denominator is the vector being projected onto.

Key insight. The identity that unlocks the hard half of this exercise is $|\vec{v}|^2 = \vec{v}\cdot\vec{v}$. It converts any question about magnitudes into one about dot products, where the algebra is ordinary and the cross terms cancel. Questions 6, 9, 11 and 13 all look quite different and are all this one idea.

Question 1

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$ respectively, having $\vec{a}\cdot\vec{b} = \sqrt{6}$.

Solution. Everything needed is already in the geometric definition — rearrange it for $\cos\theta$.

$$\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3}\times 2} = \frac{\sqrt{6}}{2\sqrt{3}}$$

Simplify by writing $\sqrt{6} = \sqrt{3}\sqrt{2}$, so the $\sqrt{3}$ cancels:

$$\cos\theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta = \frac{\pi}{4}$$

Question 2

Find the angle between the vectors $\hat{i} – 2\hat{j} + 3\hat{k}$ and $3\hat{i} – 2\hat{j} + \hat{k}$.

Solution. Here the components are given, so use the computational form for the dot product and then feed it into the geometric one.

$$\vec{a}\cdot\vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$$

$$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}, \qquad |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14}$$

$$\cos\theta = \frac{10}{\sqrt{14}\times\sqrt{14}} = \frac{10}{14} = \frac{5}{7}$$

Question 3

Find the projection of the vector $\hat{i} – \hat{j}$ on the vector $\hat{i} + \hat{j}$.

Solution. Projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$.

$$\vec{a}\cdot\vec{b} = (1)(1) + (-1)(1) = 0$$

The numerator vanishes, so the projection is zero — and the reason is worth seeing rather than just computing: a zero dot product means the two vectors are perpendicular, and a vector casts no shadow on a direction at right angles to it.

Question 4

Find the projection of the vector $\hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $7\hat{i} – \hat{j} + 8\hat{k}$.

Solution. Same formula, with the second vector supplying the magnitude in the denominator.

$$\vec{a}\cdot\vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 – 3 + 56 = 60$$

$$|\vec{b}| = \sqrt{49 + 1 + 64} = \sqrt{114}$$

$$\text{Projection} = \frac{60}{\sqrt{114}}$$

Question 5

Show that each of the given three vectors is a unit vector:

$$\frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}), \quad \frac{1}{7}(3\hat{i} – 6\hat{j} + 2\hat{k}), \quad \frac{1}{7}(6\hat{i} + 2\hat{j} – 3\hat{k})$$

Also, show that they are mutually perpendicular to each other.

Solution. Call them $\vec{a}$, $\vec{b}$ and $\vec{c}$. Each has the same three numbers $2, 3, 6$ in some order and with some signs, so each magnitude is the same calculation:

$$|\vec{a}| = \frac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7}\sqrt{4 + 9 + 36} = \frac{\sqrt{49}}{7} = 1$$

$$|\vec{b}| = \frac{1}{7}\sqrt{9 + 36 + 4} = 1, \qquad |\vec{c}| = \frac{1}{7}\sqrt{36 + 4 + 9} = 1$$

All three are unit vectors. For perpendicularity, take the three dot products. The factor $\tfrac{1}{7}\times\tfrac{1}{7}$ is positive and cannot change whether the result is zero, so it can be ignored.

$$\vec{a}\cdot\vec{b} \propto (2)(3) + (3)(-6) + (6)(2) = 6 – 18 + 12 = 0$$

$$\vec{b}\cdot\vec{c} \propto (3)(6) + (-6)(2) + (2)(-3) = 18 – 12 – 6 = 0$$

$$\vec{c}\cdot\vec{a} \propto (6)(2) + (2)(3) + (-3)(6) = 12 + 6 – 18 = 0$$

Question 6

Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a} + \vec{b})\cdot(\vec{a} – \vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.

Solution. Expand the product. Because the dot product is commutative, the two cross terms $\vec{a}\cdot\vec{b}$ and $-\vec{b}\cdot\vec{a}$ cancel exactly:

$$(\vec{a} + \vec{b})\cdot(\vec{a} – \vec{b}) = \vec{a}\cdot\vec{a} – \vec{b}\cdot\vec{b} = |\vec{a}|^2 – |\vec{b}|^2$$

So $|\vec{a}|^2 – |\vec{b}|^2 = 8$. Substituting $|\vec{a}| = 8|\vec{b}|$:

$$64|\vec{b}|^2 – |\vec{b}|^2 = 8 \quad\Rightarrow\quad 63|\vec{b}|^2 = 8 \quad\Rightarrow\quad |\vec{b}|^2 = \frac{8}{63}$$

$$|\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}}, \qquad |\vec{a}| = 8|\vec{b}| = \frac{16\sqrt{2}}{3\sqrt{7}}$$

Question 7

Evaluate the product $(3\vec{a} – 5\vec{b})\cdot(2\vec{a} + 7\vec{b})$.

Solution. Expand as ordinary algebra, then replace $\vec{a}\cdot\vec{a}$ by $|\vec{a}|^2$ and collect the two cross terms, which do not cancel this time.

$$= 6(\vec{a}\cdot\vec{a}) + 21(\vec{a}\cdot\vec{b}) – 10(\vec{b}\cdot\vec{a}) – 35(\vec{b}\cdot\vec{b})$$

Since $\vec{b}\cdot\vec{a} = \vec{a}\cdot\vec{b}$, the middle terms combine to $11(\vec{a}\cdot\vec{b})$.

$$= 6|\vec{a}|^2 + 11\,\vec{a}\cdot\vec{b} – 35|\vec{b}|^2$$

Question 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 $\tfrac{1}{2}$.

Solution. “Same magnitude” means $|\vec{a}| = |\vec{b}|$, so the geometric definition has only one unknown in it.

$$\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos 60° = |\vec{a}|^2\times\frac{1}{2}$$

Setting this equal to $\tfrac{1}{2}$:

$$\frac{|\vec{a}|^2}{2} = \frac{1}{2} \quad\Rightarrow\quad |\vec{a}|^2 = 1 \quad\Rightarrow\quad |\vec{a}| = 1$$

A magnitude cannot be negative, so the negative root is discarded.

Question 9

Find $|\vec{x}|$, if for a unit vector $\vec{a}$, $(\vec{x} – \vec{a})\cdot(\vec{x} + \vec{a}) = 12$.

Solution. The same difference-of-squares expansion as question 6.

$$(\vec{x} – \vec{a})\cdot(\vec{x} + \vec{a}) = |\vec{x}|^2 – |\vec{a}|^2$$

$\vec{a}$ is a unit vector, so $|\vec{a}| = 1$ and $|\vec{a}|^2 = 1$.

$$|\vec{x}|^2 – 1 = 12 \quad\Rightarrow\quad |\vec{x}|^2 = 13 \quad\Rightarrow\quad |\vec{x}| = \sqrt{13}$$

Question 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 $\lambda$.

Solution. Perpendicular means the dot product is zero — that single fact turns the condition into an equation in $\lambda$.

$$\vec{a} + \lambda\vec{b} = (2 – \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}$$

Note that $\vec{c} = 3\hat{i} + \hat{j} + 0\hat{k}$, so the $\hat{k}$ component contributes nothing:

$$(\vec{a} + \lambda\vec{b})\cdot\vec{c} = 3(2 – \lambda) + 1(2 + 2\lambda) + 0 = 0$$

$$6 – 3\lambda + 2 + 2\lambda = 0 \quad\Rightarrow\quad 8 – \lambda = 0$$

Question 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 $\vec{b}$.

Solution. Perpendicular means the dot product is zero, so expand it. Note that $|\vec{a}|$ and $|\vec{b}|$ are ordinary numbers and come straight out.

$$\left(|\vec{a}|\vec{b} + |\vec{b}|\vec{a}\right)\cdot\left(|\vec{a}|\vec{b} – |\vec{b}|\vec{a}\right)$$

$$= |\vec{a}|^2(\vec{b}\cdot\vec{b}) – |\vec{a}||\vec{b}|(\vec{b}\cdot\vec{a}) + |\vec{b}||\vec{a}|(\vec{a}\cdot\vec{b}) – |\vec{b}|^2(\vec{a}\cdot\vec{a})$$

The two middle terms are equal and opposite, so they cancel. Replacing $\vec{b}\cdot\vec{b}$ with $|\vec{b}|^2$ and $\vec{a}\cdot\vec{a}$ with $|\vec{a}|^2$:

$$= |\vec{a}|^2|\vec{b}|^2 – |\vec{b}|^2|\vec{a}|^2 = 0$$

Question 12

If $\vec{a}\cdot\vec{a} = 0$ and $\vec{a}\cdot\vec{b} = 0$, then what can be concluded about the vector $\vec{b}$?

Solution. Start with the first condition, which is more restrictive than it looks.

$$\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 0 \quad\Rightarrow\quad |\vec{a}| = 0 \quad\Rightarrow\quad \vec{a} = \vec{0}$$

Once $\vec{a}$ is the zero vector, the second condition $\vec{a}\cdot\vec{b} = 0$ is satisfied automatically, whatever $\vec{b}$ is — the zero vector has a zero dot product with everything. So the second condition tells us nothing at all about $\vec{b}$.

Question 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 $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}$.

Solution. The quantity wanted is a sum of dot products, and the condition given is about a vector being zero. The bridge between them is squaring the magnitude — take the dot product of the sum with itself.

$$|\vec{a} + \vec{b} + \vec{c}|^2 = 0$$

Expanding, exactly as $(x+y+z)^2$ expands:

$$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$$

All three are unit vectors, so each squared magnitude is $1$:

$$3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$$

Question 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.

Solution. The forward statement is immediate: the zero vector has zero components, so every term of the dot product is zero.

The converse would claim that $\vec{a}\cdot\vec{b} = 0$ forces one of them to be zero. A single counterexample defeats it. Take

$$\vec{a} = \hat{i}, \qquad \vec{b} = \hat{j}$$

Then $\vec{a}\cdot\vec{b} = (1)(0) + (0)(1) + (0)(0) = 0$, yet $|\vec{a}| = |\vec{b}| = 1$, so neither is the zero vector.

The reason is geometric: $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$ vanishes when $\cos\theta = 0$, that is when the vectors are perpendicular — which has nothing to do with either being zero.

Question 15

If the vertices A, B, C of a triangle ABC are $(1, 2, 3)$, $(-1, 0, 0)$, $(0, 1, 2)$ respectively, then find $\angle\mathrm{ABC}$.

Solution. $\angle\mathrm{ABC}$ is the angle at B, so it is the angle between the two vectors that leave B — namely $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$. Both must be built with B as the initial point; using $\overrightarrow{\mathrm{AB}}$ instead gives the supplementary angle.

$$\overrightarrow{\mathrm{BA}} = (1 – (-1))\hat{i} + (2 – 0)\hat{j} + (3 – 0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}$$

$$\overrightarrow{\mathrm{BC}} = (0 – (-1))\hat{i} + (1 – 0)\hat{j} + (2 – 0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$$

$$\overrightarrow{\mathrm{BA}}\cdot\overrightarrow{\mathrm{BC}} = 2 + 2 + 6 = 10$$

$$|\overrightarrow{\mathrm{BA}}| = \sqrt{4 + 4 + 9} = \sqrt{17}, \qquad |\overrightarrow{\mathrm{BC}}| = \sqrt{1 + 1 + 4} = \sqrt{6}$$

$$\cos(\angle\mathrm{ABC}) = \frac{10}{\sqrt{17}\sqrt{6}} = \frac{10}{\sqrt{102}}$$

Question 16

Show that the points $\mathrm{A}(1, 2, 7)$, $\mathrm{B}(2, 6, 3)$ and $\mathrm{C}(3, 10, -1)$ are collinear.

Solution. Three points are collinear when the longest of the three distances between them equals the sum of the other two — there is no “detour”.

$$\overrightarrow{\mathrm{AB}} = \hat{i} + 4\hat{j} – 4\hat{k}, \qquad |\overrightarrow{\mathrm{AB}}| = \sqrt{1 + 16 + 16} = \sqrt{33}$$

$$\overrightarrow{\mathrm{BC}} = \hat{i} + 4\hat{j} – 4\hat{k}, \qquad |\overrightarrow{\mathrm{BC}}| = \sqrt{33}$$

$$\overrightarrow{\mathrm{AC}} = 2\hat{i} + 8\hat{j} – 8\hat{k}, \qquad |\overrightarrow{\mathrm{AC}}| = \sqrt{4 + 64 + 64} = \sqrt{132} = 2\sqrt{33}$$

$$|\overrightarrow{\mathrm{AB}}| + |\overrightarrow{\mathrm{BC}}| = \sqrt{33} + \sqrt{33} = 2\sqrt{33} = |\overrightarrow{\mathrm{AC}}|$$

so A, B and C are collinear. In fact $\overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{BC}}$ exactly, which says something stronger: B is the midpoint of AC.

Question 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.

Solution. Treat the three vectors as position vectors of A, B and C, find the three sides, and compare squared lengths. Working with squares keeps every number an integer.

$$\overrightarrow{\mathrm{AB}} = -\hat{i} – 2\hat{j} – 6\hat{k}, \qquad |\overrightarrow{\mathrm{AB}}|^2 = 1 + 4 + 36 = 41$$

$$\overrightarrow{\mathrm{BC}} = 2\hat{i} – \hat{j} + \hat{k}, \qquad |\overrightarrow{\mathrm{BC}}|^2 = 4 + 1 + 1 = 6$$

$$\overrightarrow{\mathrm{CA}} = -\hat{i} + 3\hat{j} + 5\hat{k}, \qquad |\overrightarrow{\mathrm{CA}}|^2 = 1 + 9 + 25 = 35$$

The two smaller squares are $6$ and $35$, and

$$|\overrightarrow{\mathrm{BC}}|^2 + |\overrightarrow{\mathrm{CA}}|^2 = 6 + 35 = 41 = |\overrightarrow{\mathrm{AB}}|^2$$

By the converse of Pythagoras’ theorem the triangle is right angled, at the vertex where the two shorter sides meet — that is at C, opposite the longest side AB.

Since this exercise is about the dot product, the check can also be done directly: the angle at C is between $\overrightarrow{\mathrm{CB}} = -2\hat{i} + \hat{j} – \hat{k}$ and $\overrightarrow{\mathrm{CA}} = -\hat{i} + 3\hat{j} + 5\hat{k}$, and

$$\overrightarrow{\mathrm{CB}}\cdot\overrightarrow{\mathrm{CA}} = 2 + 3 – 5 = 0$$

Question 18

If $\vec{a}$ is a nonzero vector of magnitude ‘$a$’ and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is unit vector if

Solution. Scaling a vector by $\lambda$ multiplies its length by $|\lambda|$ — the modulus, because a negative $\lambda$ reverses the direction but cannot make a length negative.

$$|\lambda\vec{a}| = |\lambda|\,|\vec{a}| = |\lambda|\,a$$

For this to be a unit vector, set it equal to $1$:

$$|\lambda|\,a = 1 \quad\Rightarrow\quad a = \frac{1}{|\lambda|}$$

Options (A) and (B) fail because they say nothing about $a$, and $\vec{a}$ is only a unit vector to begin with if $a = 1$. Option (C) is the relation upside down.

Common mistakes

• Dividing by the wrong magnitude in a projection. The projection of $\vec{a}$ on $\vec{b}$ divides by $|\vec{b}|$, the vector being projected onto. Questions 3 and 4 both hinge on getting this the right way round.
• Cancelling the cross terms when they do not cancel. In $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$ they genuinely do, giving $|\vec{a}|^2 – |\vec{b}|^2$ (questions 6 and 9). In question 7 they combine to $11\,\vec{a}\cdot\vec{b}$ instead. Expand properly rather than assuming.
• Forgetting $|\vec{v}|^2 = \vec{v}\cdot\vec{v}$. Question 13 is unreachable without it: the condition is about a vector, the answer is about dot products, and squaring the magnitude is the only bridge.
• Building the wrong vectors for an angle at a vertex. $\angle\mathrm{ABC}$ is between $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$, both starting at B. Using $\overrightarrow{\mathrm{AB}}$ gives the supplementary angle and a sign error in the cosine. Question 15.
• Concluding a vector is zero from $\vec{a}\cdot\vec{b} = 0$. It means perpendicular, not zero — that is the whole point of question 14. But note that $\vec{a}\cdot\vec{a} = 0$ does force $\vec{a} = \vec{0}$, which is what question 12 turns on. The two look similar and behave differently.
• Dropping the modulus in question 18. $|\lambda\vec{a}| = |\lambda|\,a$, not $\lambda a$; a negative scalar still lengthens by a positive factor.

Practise next

• Exercise 10.4 — the vector (cross) product, which answers the questions the dot product cannot: area and perpendicular direction.
• Miscellaneous Exercise on Chapter 10 — mixed questions drawing on the whole chapter.