NCERT Class 12 Mathematics — Vector Algebra, Exercise 10.4. All 12 questions solved.

Exercise 10.4 is the vector product, or cross product. Where the dot product of Exercise 10.3 returned a number, this returns a vector — one perpendicular to both of the originals, with magnitude

$$|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta$$

In practice it is computed as a determinant:

$$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

Two geometric readings account for most of the exercise. That magnitude is the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as adjacent sides, so half of it is the area of the triangle. And dividing $\vec{a}\times\vec{b}$ by its own magnitude gives a unit vector perpendicular to both.

Key insight. Hold the two products apart in your head, because they answer opposite questions. $\vec{a}\cdot\vec{b} = 0$ means perpendicular; $\vec{a}\times\vec{b} = \vec{0}$ means parallel. Question 6 exists purely to make you notice that both at once is impossible unless one of the vectors is zero.

Remember also that the cross product is not commutative: $\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b})$, and $\vec{v}\times\vec{v} = \vec{0}$. Question 4 turns on both.

Question 1

Find $|\vec{a}\times\vec{b}|$, if $\vec{a} = \hat{i} – 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k}$.

Solution. Set up the determinant and expand along the top row.

$$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}$$

$$= \hat{i}\big[(-7)(2) – (7)(-2)\big] – \hat{j}\big[(1)(2) – (7)(3)\big] + \hat{k}\big[(1)(-2) – (-7)(3)\big]$$

$$= \hat{i}(-14 + 14) – \hat{j}(2 – 21) + \hat{k}(-2 + 21) = 0\hat{i} + 19\hat{j} + 19\hat{k}$$

Note the minus sign in front of the $\hat{j}$ term — it belongs to the determinant expansion, not to the arithmetic, and dropping it is the single most common slip in this exercise.

$$|\vec{a}\times\vec{b}| = \sqrt{0 + 361 + 361} = \sqrt{722} = 19\sqrt{2}$$

Question 2

Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b}$ and $\vec{a} – \vec{b}$, where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} – 2\hat{k}$.

Solution. A vector perpendicular to two given vectors is their cross product, so form the two vectors first and cross those — not $\vec{a}$ and $\vec{b}$ themselves.

$$\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j}, \qquad \vec{a} – \vec{b} = 2\hat{i} + 4\hat{k}$$

$$(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16 – 0) – \hat{j}(16 – 0) + \hat{k}(0 – 8)$$

$$= 16\hat{i} – 16\hat{j} – 8\hat{k}$$

$$\big|(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})\big| = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$$

Dividing by 24 gives the unit vector. There are two such unit vectors, pointing in opposite directions, so the answer carries a $\pm$:

$$\pm\left(\frac{2}{3}\hat{i} – \frac{2}{3}\hat{j} – \frac{1}{3}\hat{k}\right)$$

Question 3

If a unit vector $\vec{a}$ makes angles $\dfrac{\pi}{3}$ with $\hat{i}$, $\dfrac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\vec{a}$.

Solution. The angles a unit vector makes with the axes are exactly its direction cosines, and those satisfy

$$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

Substituting $\cos\tfrac{\pi}{3} = \tfrac{1}{2}$ and $\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$:

$$\frac{1}{4} + \frac{1}{2} + \cos^2\theta = 1 \quad\Rightarrow\quad \cos^2\theta = \frac{1}{4}$$

So $\cos\theta = \pm\tfrac{1}{2}$. The question says $\theta$ is acute, which forces the positive root — this is why that word is in the question.

$$\cos\theta = \frac{1}{2} \quad\Rightarrow\quad \theta = \frac{\pi}{3}$$

The components of a unit vector are its direction cosines:

$$\vec{a} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k}$$

Question 4

Show that $(\vec{a} – \vec{b})\times(\vec{a} + \vec{b}) = 2(\vec{a}\times\vec{b})$.

Solution. Expand as ordinary algebra, but keep the order of each product, because the cross product is not commutative.

$$(\vec{a} – \vec{b})\times(\vec{a} + \vec{b}) = \vec{a}\times\vec{a} + \vec{a}\times\vec{b} – \vec{b}\times\vec{a} – \vec{b}\times\vec{b}$$

Now apply the two rules. Any vector crossed with itself is $\vec{0}$, since the angle between them is zero and $\sin 0 = 0$, so the first and last terms vanish. And $\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b})$, so $-\vec{b}\times\vec{a}$ becomes $+\vec{a}\times\vec{b}$:

$$= \vec{0} + \vec{a}\times\vec{b} + \vec{a}\times\vec{b} – \vec{0} = 2(\vec{a}\times\vec{b})$$

Compare this with question 6 of Exercise 10.3, where the same bracket under a dot product gave $|\vec{a}|^2 – |\vec{b}|^2$: there the cross terms cancelled, here they double.

Question 5

Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k})\times(\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.

Solution. A cross product is the zero vector precisely when the two vectors are parallel, and parallel vectors have proportional components. That is far quicker than expanding the determinant and solving three equations.

$$\frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu}$$

The first ratio is $2$, so:

$$\frac{6}{\lambda} = 2 \Rightarrow \lambda = 3, \qquad \frac{27}{\mu} = 2 \Rightarrow \mu = \frac{27}{2}$$

Question 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 $\vec{b}$?

Solution. Take the two conditions in turn, assuming for the moment that both vectors are non-zero.

$\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0$ forces $\cos\theta = 0$, so $\theta = 90°$ — the vectors are perpendicular.

$\vec{a}\times\vec{b} = \vec{0}$ gives $|\vec{a}||\vec{b}|\sin\theta = 0$, so $\sin\theta = 0$ and $\theta = 0°$ or $180°$ — the vectors are parallel.

No angle is both $90°$ and $0°$, so the assumption fails: at least one of the vectors must be the zero vector, which satisfies both conditions trivially.

Question 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 $\vec{a}\times(\vec{b} + \vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}$.

Solution. Work out each side as a determinant and compare. On the left, $\vec{b} + \vec{c}$ has components $b_1 + c_1$, $b_2 + c_2$, $b_3 + c_3$:

$$\vec{a}\times(\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 + c_1 & b_2 + c_2 & b_3 + c_3 \end{vmatrix}$$

Taking just the $\hat{i}$ component:

$$a_2(b_3 + c_3) – a_3(b_2 + c_2) = (a_2b_3 – a_3b_2) + (a_2c_3 – a_3c_2)$$

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 is a difference of products that is linear in the second row. Hence the two sides agree component by component.

Question 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.

Solution. The forward statement holds because every entry in the determinant’s second or third row is zero.

The converse — that $\vec{a}\times\vec{b} = \vec{0}$ forces one of them to be zero — is not true, and one counterexample settles it. Take

$$\vec{a} = \hat{i}, \qquad \vec{b} = 2\hat{i}$$

Both are non-zero, but they are parallel, so $\sin\theta = 0$ and $\vec{a}\times\vec{b} = \vec{0}$.

This is the mirror image of question 14 in Exercise 10.3: there a vanishing dot product meant perpendicular, here a vanishing cross product means parallel. Neither implies a zero vector.

Question 9

Find the area of the triangle with vertices $\mathrm{A}(1, 1, 2)$, $\mathrm{B}(2, 3, 5)$ and $\mathrm{C}(1, 5, 5)$.

Solution. The area of a triangle on two adjacent sides is half the area of the parallelogram they span, so build two sides from the same vertex.

$$\overrightarrow{\mathrm{AB}} = \hat{i} + 2\hat{j} + 3\hat{k}, \qquad \overrightarrow{\mathrm{AC}} = 0\hat{i} + 4\hat{j} + 3\hat{k}$$

$$\overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = \hat{i}(6 – 12) – \hat{j}(3 – 0) + \hat{k}(4 – 0)$$

$$= -6\hat{i} – 3\hat{j} + 4\hat{k}$$

$$\left|\overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AC}}\right| = \sqrt{36 + 9 + 16} = \sqrt{61}$$

$$\text{Area} = \frac{1}{2}\sqrt{61}$$

Question 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 $\vec{b} = 2\hat{i} – 7\hat{j} + \hat{k}$.

Solution. For a parallelogram the area is the full magnitude of the cross product — no halving this time.

$$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = \hat{i}(-1 + 21) – \hat{j}(1 – 6) + \hat{k}(-7 + 2)$$

$$= 20\hat{i} + 5\hat{j} – 5\hat{k}$$

$$\text{Area} = |\vec{a}\times\vec{b}| = \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2}$$

Question 11

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \dfrac{\sqrt{2}}{3}$, then $\vec{a}\times\vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is

Solution. “Is a unit vector” means its magnitude is $1$, so use the geometric form of the cross product’s magnitude.

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

Setting this equal to $1$:

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

Question 12

Area of a rectangle having vertices A, B, C and D with position vectors $-\hat{i} + \tfrac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \tfrac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} – \tfrac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} – \tfrac{1}{2}\hat{j} + 4\hat{k}$ respectively is

Solution. Every position vector has the same $\hat{k}$ component of $4$, so all four points lie in the plane $z = 4$ and the $\hat{k}$ terms will cancel out of every side. Take two adjacent sides:

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

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

These are perpendicular, as they must be for a rectangle, with lengths $2$ and $1$. The area is the magnitude of their cross product:

$$\overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{BC}} = (2\hat{i})\times(-\hat{j}) = -2(\hat{i}\times\hat{j}) = -2\hat{k}$$

$$\text{Area} = |-2\hat{k}| = 2$$

Common mistakes

• Losing the minus sign on the $\hat{j}$ term. The determinant expansion is $+\hat{i}(\ldots) – \hat{j}(\ldots) + \hat{k}(\ldots)$. It costs the answer every time and appears in questions 1, 2, 9 and 10.
• Crossing the wrong pair of vectors. Question 2 asks for a vector perpendicular to $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, so those are what get crossed — not $\vec{a}$ and $\vec{b}$.
• Forgetting the $\pm$ on a perpendicular unit vector. There are always two, one on each side of the plane. Question 2.
• Halving when you should not, or not halving when you should. The cross product’s magnitude is the parallelogram area (question 10); a triangle takes half of it (question 9).
• Treating the cross product as commutative. $\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b})$, and question 4 works only because of that sign.
• Reading a zero cross product as “one of them is zero”. It means parallel. Question 8 is exactly this point, and question 6 shows what it takes to force a zero vector: a zero dot product as well.
• Ignoring the word “acute” in question 3. $\cos^2\theta = \tfrac14$ has two roots, and only that word picks one.

Practise next

• Miscellaneous Exercise on Chapter 10 — mixed questions using the dot and cross products together.
• Chapter 11, Three Dimensional Geometry — where these products become the standard tools for distances and angles between lines and planes.