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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

QuestionEASY

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 \hat{n} is a unit vector such that \vec{u}\cdot \hat{n}=0 and \vec{v}\cdot \hat{n}=0, then \left |\vec{w}\cdot \widehat{n} \right |  is equal to:

View Solution

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

View Solution

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

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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

View Solution

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 :

View Solution

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

Study other Related Concepts

Cross product Current Topic

3D Geometry
Vectors and Scalars
Vectors Joining Two Points
Types Of Vectors
Direction Cosines & Direction Ratios Of A Line
Vector Addition and Subtraction
Multiplication Of Vectors And Scalar Quantity
Section Formula
Linear Combination of Vectors
Dot Product Of Two Vectors

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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

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