# Cross Product of Vectors Formula [ Vector Product ]

## Cross Product of Vectors Formula :

Let $$\vec{a}$$ & $$\vec{b}$$ are two vectors & $$\theta$$ is the angle between them, then cross product of vectors formula is,

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

where $$\hat{n}$$ is the unit vector perpendicular to both $$\vec{a}$$ & $$\vec{b}$$.

## Properties of Vector Cross Product :

(i) $$\vec{a}$$ $$\times$$ $$\vec{b}$$ = $$\vec{0}$$ $$\iff$$ $$\vec{a}$$ & $$\vec{b}$$ are parallel(Collinear) ($$\vec{a}$$ $$\ne$$ 0, $$\vec{b}$$ $$\ne$$ 0) i.e. $$\vec{a}$$ = K$$\vec{b}$$, where K is a scalar.

(ii) $$\vec{a}$$ $$\times$$ $$\vec{b}$$ $$\ne$$ $$\vec{b}$$ $$\times$$ $$\vec{a}$$  (not commutative)

(iii) m($$\vec{a}$$) $$\times$$ $$\vec{b}$$ = $$\vec{a}$$ $$\times$$ (m$$\vec{b}$$) = m($$\vec{a}$$ $$\times$$ $$\vec{b}$$) where m is a scalar.

(iv) $$\vec{a}$$ $$\times$$ ($$\vec{b}$$ + $$\vec{c}$$)  (distributive over addition)

(v)  $$\hat{i}$$ $$\times$$ $$\hat{i}$$ = $$\hat{j}$$ $$\times$$ $$\hat{j}$$ = $$\hat{k}$$ $$\times$$ $$\hat{k}$$ = 0

(vi) $$\hat{i}$$ $$\times$$ $$\hat{j}$$ = $$\hat{k}$$, $$\hat{j}$$ $$\times$$ $$\hat{k}$$ = $$\hat{i}$$, $$\hat{k}$$ $$\times$$ $$\hat{i}$$ = $$\hat{j}$$

(vii) If $$\vec{a}$$ = $$a_1\hat{i}$$ + $$a_2\hat{j}$$ + $$a_3\hat{k}$$  $$\vec{b}$$ = $$b_1\hat{i}$$ + $$b_2\hat{j}$$ + $$b_3\hat{k}$$, then $$\vec{a}$$ $$\times$$ $$\vec{b}$$ = $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{vmatrix}$$

Example : Find $$\vec{a}$$ $$\times$$ $$\vec{b}$$, if $$\vec{a}$$ = $$2\hat{i} +\hat{k}$$ and $$\vec{b}$$ = $$\hat{i} +\hat{j} + \hat{k}$$

Solution : We have, $$\vec{a}$$ = $$2\hat{i} +\hat{k}$$ and $$\vec{b}$$ = $$\hat{i} +\hat{j} + \hat{k}$$

$$\therefore$$   $$\vec{a}\times\vec{b}$$ = $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \\ \end{vmatrix}$$ = $$-1\hat{i} – 1\hat{j} + 2\hat{k}$$

## Vectors Normal to the Plane of Two Given Vectors :

Let $$\vec{a}$$ & $$\vec{b}$$ be two non-zero, non-parallel vectors and let $$\theta$$ be the angle between them.

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

where $$\hat{n}$$ is the unit vector perpendicular to plane of $$\vec{a}$$ & $$\vec{b}$$ such that $$\vec{a}$$, $$\vec{b}$$, $$\hat{n}$$ form a right-handed system.

$$\therefore$$ $$\vec{a}$$ $$\times$$ $$\vec{b}$$ = | $$\vec{a}$$ $$\times$$ $$\vec{b}$$|$$\hat{n}$$

$$\implies$$ $$\hat{n}$$ = $$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$$

Thus, $$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$$ is a unit vector perpendicular to the plane of $$\vec{a}$$ and $$\vec{b}$$.

Note that -$$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$$ is also a unit vector perpendicular to the plane of $$\vec{a}$$ and $$\vec{b}$$.

Vector of magnitude ‘$$\lambda$$’ normal to the plane of $$\vec{a}$$ and $$\vec{b}$$ are given by $$\pm$$$$\lambda(\vec{a}\times\vec{b})\over |\vec{a}\times\vec{b}|$$.

Example : Find a unit vector perpendicular to both the vectors $$\hat{i} -2\hat{j} + 3\hat{k}$$ and $$\hat{i} + 2\hat{j} – \hat{k}$$.

Solution : Let $$\vec{a}$$ = $$\hat{i} -2\hat{j} + 3\hat{k}$$ and $$\vec{b}$$ = $$\hat{i} + 2\hat{j} – \hat{k}$$

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

= $$-4\hat{i} + 4\hat{j} + 4\hat{k}$$

$$\therefore$$ | $$\vec{a}$$ $$\times$$ $$\vec{b}$$| = $$\sqrt{(-4)^2+4^2+4^2}$$ = 4$$\sqrt{3}$$

Hence, a unit vector perpendicular to vectors $$\vec{a}$$ and $$\vec{b}$$ is given by

$$\hat{n}$$ = $$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$$ = $$-4\hat{i} + 4\hat{j} + 4\hat{k}\over 4\sqrt{3}$$ = $$1\over \sqrt{3}$$($$-\hat{i}+\hat{j}+\hat{k}$$).

Hope you learnt cross product of vectors formula, learn more concepts of vectors and practice more questions to get ahead in the competition. Good luck!