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# Cross product - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Vector (or Cross) Product of Two Vectors is considered one the most difficult concept.

• Vector Product in Terms of Components is considered one of the most asked concept.

• 55 Questions around this concept.

## Solve by difficulty

Let $\dpi{100} \vec{u}=\hat{i}+\hat{j},\; \; \vec{v}=\hat{i}-\hat{j}$ and $\vec{w}=\hat{i}+2\hat{j}+3\hat{k}$. If $\dpi{100} \hat{n}$ is a unit vector such that $\dpi{100} \vec{u}\cdot \hat{n}=0$ and $\dpi{100} \vec{v}\cdot \hat{n}=0,$ then $\left |\vec{w}\cdot \widehat{n} \right |$  is equal to:

If $\dpi{100} \left | \vec{a} \right |=4,\left | \vec{b} \right |=2,$  and the angle between $\dpi{100} \vec{a}\; \; and \; \; \vec{b} \; \; is \; \; \frac{\pi }{6}\; then\; \; (\vec{a}\times \vec{b} )^{2}$ is equal to

$\dpi{100} \vec{a}=3\hat{i}-5\hat{j}\;\; and\;\; \vec{b}=6\hat{i}+3\hat{j}$  are two vectors and $\dpi{100} \vec{c}$ is a vector such that $\dpi{100} \vec{c}=\vec{a}\times \vec{b}$ then $\dpi{100} \left | \vec{a} \right |:\left |\vec{b} \right |:\left | \vec{c} \right |=$$| \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} +\vec{c} \cdot \vec{a} |$

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$. Then $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is equal to

Let $\overrightarrow{\mathrm{a}}=-5 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{\mathrm{i}}) \times \hat{\mathrm{i}}) \times \hat{\mathrm{i}}$. Then $\overrightarrow{\mathrm{c}} \cdot(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ is equal to :

## Concepts Covered - 2

Vector (or Cross) Product of Two Vectors

The vector product of two nonzero vectors $\vec{\mathbf a}$ and $\vec{\mathbf b}$, is denoted by $\vec{\mathbf a} \times \vec{\mathbf b}$  and defined as,

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

where Ө is the angle between $\vec{\mathbf a}$ and $\vec{\mathbf b}$, 0 ≤ θ ≤ π and $\mathbf{\hat n}$ is a unit vector perpendicular to both $\vec{\mathbf a}$ and $\vec{\mathbf b}$, such that $\vec{\mathbf a}$$\vec{\mathbf b}$ and $\mathbf{\hat n}$ form a right hand system.

Observe that, the direction of $\vec{\mathbf a} \times \vec{\mathbf b}$  is opposite to that of $\vec{\mathbf b} \times \vec{\mathbf a}$ as shown in the figure.

i.e.  $\vec{\mathbf a} \times \vec{\mathbf b} =-\vec{\mathbf a} \times \vec{\mathbf b}$

So the vector product is not commutative.

Properties of Vector Product
$\\1.\;\;\;\;\vec a\times \vec b\;\text{ is a vector.}\\2.\;\;\;\text { Let } \vec{a} \text { and } \vec{b} \text { be two nonzero vectors. Then } \vec{a} \times \vec{b}=\overrightarrow{0} \text { if and only if } \vec{a} \text { and } \vec{b}\\\mathrm{\;\;\;\;\;}\text{\;\;are parallel (or collinear) to each other, i.e.,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \vec{a} \times \vec{b}=\overrightarrow{0} \Leftrightarrow \vec{a} \parallel \vec{b}\\\mathrm{\;\;\;\;\;\;}\text { In particular, } \vec{a} \times \vec{a}=\overrightarrow{0} \; \text { and } \vec{a} \times(-\vec{a})=\overrightarrow{0}, \text { since in the first situation, }\\\mathrm{\;\;\;\;\;\;\;}\theta=0\;\text { and in the second one, } \theta=\pi.\\3.\mathrm{\;\;\;\;}\text{If }\theta=\frac{\pi}{2},\;\;\text{then }\;\vec a\times \vec b=|\vec a||\vec b|.$

$\\4.\;\;\;\;\text{From the property 2. and 3.}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;} \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;}\hat{i} \times \hat{j}=\hat{k}, \quad \hat{j} \times \hat{k}=\hat{i}, \quad \hat{k} \times \hat{i}=\hat{j}\\\text{\;\;\;\;\;\;\;\;\;\;and also,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;} \hat{j} \times \hat{i}=-\hat{k}, \quad \hat{k} \times \hat{j}=-\hat{i} \text { and } \quad \hat{i} \times \hat{k}=-\hat{j}$

$\\5.\;\;\;\;\text { Vector product is not associative, }\\\mathrm{\;\;\;\;\;\;\;\;\;i.e.\;\;\;\;}\vec a\times \vec b\neq \vec b\times \vec a\\6.\;\;\;\;\text { If } \vec a \text{ and } \vec b \text{ are two vectors and } m \text { is a scalar, then }\\\mathrm{\;\;\;\;\;\;\;\;}m \vec{a} \times \vec{b}=m(\vec{a} \times \vec{b})=\vec{a} \times m \vec{b}\\7.\;\;\;\;\text { If } \vec a \text{ and } \vec b \text{ are two vectors and } m,\;n \text { are scalars, then }\\\mathrm{\;\;\;\;\;\;\;\;}m \vec a \times n \vec b=m n(\vec a \times\vec b)=m(n\vec a \times\vec b)=n(m \vec a \times \vec b)\\8.\;\;\;\; \text { For any three vectors}\;\;\vec a,\;\vec b \;\text{and}\;\vec c,\text{ we have }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}(i)\;\;\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}(ii)\;\vec{a} \times(\vec{b}-\vec{c})=\vec{a} \times \vec{b}-\vec{a} \times \vec{c}$

Vector Product in Terms of Components

$\\ \text {If } \vec {\mathbf a}=a_{1} \hat{\mathbf{i}}+a_{2} \hat{\mathbf{j}}+a_{3} \hat{\mathbf{k}} \;\text { and }\; \vec {\mathbf{b}}=b_{1} \hat{\mathbf{i}}+b_{2} \hat{\mathbf{j}}+b_{3} \hat{\mathbf{k}}, \text { then their cross product given by}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \vec{a} \times \vec{b}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_{1}} & {a_{2}} & {a_{3}} \\ {b_{1}} & {b_{2}} & {b_{3}}\end{array}\right|$

Proof:

$\\\vec{a} \times \vec{b}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)\\\mathrm{\;\;\;\;\;\;\;\;}=a_{1} b_{1}(\hat{i} \times \hat{i})+a_{1} b_{2}(\hat{i} \times \hat{j})+a_{1} b_{3}(\hat{i} \times \hat{k})+a_{2} b_{1}(\hat{j} \times \hat{i})\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}+a_{2} b_{2}(\hat{j} \times \hat{j})+a_{2} b_{3}(\hat{j} \times \hat{k})+a_{3} b_{1}(\hat{k} \times \hat{i})+a_{3} b_{2}(\hat{k} \times \hat{j})+a_{3} b_{3}(\hat{k} \times \hat{k})\\\mathrm{\;\;\;\;\;\;\;\;} =a_{1} b_{2}(\hat{i} \times \hat{j})-a_{1} b_{3}(\hat{k} \times \hat{i})-a_{2} b_{1}(\hat{i} \times \hat{j}) \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}+ +a_{2} b_{3}(\hat{j} \times \hat{k})+a_{3} b_{1}(\hat{k} \times \hat{i})-a_{3} b_{2}(\hat{j} \times \hat{k})$

$\\\text{Using the fact that,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0\\\text{and,}\;\;\;\;\;\;\;\hat{i} \times \hat{k}=-\hat{k} \times \hat{i}, \hat{j} \times \hat{i}=-\hat{i} \times \hat{j} \text { and } \hat{k} \times \hat{j}=-\hat{j} \times \hat{k}\\\mathrm{\;\;\;\;\;\;\;\;}=a_{1} b_{2} \hat{k}-a_{1} b_{3} \hat{j}-a_{2} b_{1} \hat{k}+a_{2} b_{3} \hat{i}+a_{3} b_{1} \hat{j}-a_{3} b_{2} \hat{i}\\\text { (as }\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i} \text { and } \hat{k} \times \hat{i}=\hat{j})\\\mathrm{\;\;\;\;\;\;\;\;}=\left(a_{2} b_{3}-a_{3} b_{2}\right) \hat{i}-\left(a_{1} b_{3}-a_{3} b_{1}\right) \hat{j}+\left(a_{1} b_{2}-a_{2} b_{1}\right) \hat{k}\\\mathrm{\;\;\;\;\;\;\;\;}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_{1}} & {a_{2}} & {a_{3}} \\ {b_{1}} & {b_{2}} & {b_{3}}\end{array}\right|$

Angle between Two Vectors

If Ө is the angle between the vectors $\vec {\mathbf a}$ and $\vec{\mathbf b}$, then

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

Vector perpendicular to the plane of two given vectors

The unit vector perpendicular to the plane of $\vec {\mathbf a}$ and $\vec{\mathbf b}$ is $\frac{(\vec{\mathbf a}\times \vec{\mathbf b})}{|\vec {\mathbf a}\times\vec {\mathbf b}|} .$

Also note that $-\frac{(\vec{\mathbf a}\times \vec{\mathbf b})}{|\vec {\mathbf a}\times\vec {\mathbf b}|}$is also a unit vector perpendicular to the plane of $\vec {\mathbf a}$ and $\vec{\mathbf b}$.

Vectors of magnitude ‘λ’ perpendicular to the plane of $\vec {\mathbf a}$ and $\vec{\mathbf b}$ are given by $\pm\frac{\lambda(\vec{\mathbf a}\times \vec{\mathbf b})}{|\vec {\mathbf a}\times\vec {\mathbf b}|}$

## Study it with Videos

Vector (or Cross) Product of Two Vectors
Vector Product in Terms of Components

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## Books

### Reference Books

#### Vector (or Cross) Product of Two Vectors

Mathematics for Joint Entrance Examination JEE (Advanced) : Vectors and 3D Geometry

Page No. : 3.20

Line : 1

#### Vector Product in Terms of Components

Mathematics for Joint Entrance Examination JEE (Advanced) : Vectors and 3D Geometry

Page No. : 3.20

Line : 1