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# Dot Product Of Two Vectors - Practice Questions & MCQ

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

## Quick Facts

• Dot (Scalar) Product in Terms of Components is considered one the most difficult concept.

• Dot (Scalar) Product of Two Vectors is considered one of the most asked concept.

• 67 Questions around this concept.

## Solve by difficulty

Let $\dpi{100} \vec{a}=\hat{j}-\hat{k}\;$  and $\; \vec{c}=\hat{i}-\hat{j}-\hat{k}$.  Then the vector $\vec{b}$ satisfying $\dpi{100} \vec{a}\times \vec{b}+ \vec{c}=0\;$ and $\; \vec{a}\cdot \vec{b}=3\;$ is

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 $\dpi{100} \left | \vec{a} \right |=2,\left | \vec{b} \right |=3\: and\: \left | 2\; \vec{a}-\vec{b} \right | = 5\: then \: \left | 2\; \vec{a} +\vec{b}\right |$ equals:

$\begin{array}{l}{\text { Let } \vec{a}=2 \hat{i}+\hat{j}-2 \hat{k} \text { and } \vec{b}=\hat{i}+\hat{j}} \\ {\text { Let } \vec{c} \text { be a vector such that }|\vec{c}-\vec{a}|=3,|(\vec{a} \times \vec{b}) \times \vec{c}|=3} \\ {\text { and the angle between } \vec{c} \text { and } \vec{a} \times \vec{b} \text { is } 30^{\circ} . \text { Then } \vec{a} . \vec{c} \text { is equal to }}\end{array}$

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively

then the point (p, q) lies on a line :

Let ABCD be a parallelogram such that ,      and   be an acute angle. If   is the vector that coincides with the altitude directed from the vertex B to the side AD, then is given by

The position vectors of the vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a triangle are $2 \hat{\imath}-3 \hat{\jmath}+3 \hat{k}, 2 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}$ and $-\hat{\imath}+\hat{\jmath}+3 \hat{k}$ respectively. Let $\ell$ denotes the length of the angle bisector $\mathrm{AD}$ of $\angle \mathrm{BAC}$ where $\mathrm{D}$ is on the line segment $\mathrm{BC}$, then $2 \ell^2$ equals:

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## Concepts Covered - 2

Dot (Scalar) Product of Two Vectors

Multiplication (or product) of two vectors is defined in two ways, namely, dot (or scalar) product where the result is a scalar, and vector (or cross) product where the result is a vector. Based upon these two types of products for vectors, we have various applications in geometry, mechanics and engineering.

Dot (scalar) Product

If $\vec{\mathbf a}$ and $\vec{\mathbf b}$ are two non-zero vectors, then their scalar product (or dot product) is denoted by $\vec{\mathbf a} \cdot \vec{\mathbf b}$ and is defined as

$\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\vec{\mathbf a}||\vec{\mathbf b}| \cos \theta. \;\;\;\;\;\;\;\;(0\leq \theta\leq \pi)$ where, θ is the angle between  $\vec{\mathbf a}$ and $\vec{\mathbf b}$

Observations:

$\\\text { 1. } \;\;\;\vec{a} \cdot \vec{b} \text { is a real number. }\\\text { 2. } \;\;\;\vec{a} \cdot \vec{b}\text { is positive if } \theta \text { is acute. }\\\text { 3. } \;\;\;\vec{a} \cdot \vec{b}\text { is negative if } \theta \text { is obtuse. }\\\text { 4. } \;\;\;\vec{a} \cdot \vec{b}\text { is zero if } \theta \text { is }90^\circ.\\\text { 5. } \;\;\;\vec{a} \cdot \vec{b}\leq|\vec{a}| | \vec{b}|$

$\\\text {For any two non-zero vectors } \vec {\mathbf a} \text{ and }\vec { \mathbf b}, \text{then}\mathrm{\;\;\;}\vec {\mathbf a} \cdot\vec { \mathbf b}=0\text{ if and only if }\vec {\mathbf a} \text{ and }\vec { \mathbf b}\\\text{perpendicular to each other. i.e.}\\\vec {\mathbf a} \cdot\vec { \mathbf b}=0\Leftrightarrow \vec {\mathbf a} \perp \vec {\mathbf b} \\\text {As } \hat{\mathbf{i}}, \hat{\mathbf{j}} \text { and } \hat{\mathbf{k}} \text { are mutually perpendicular unit vectors }\text { along the coordinate axes, }\\\text{therefore,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\hat{\mathbf{i}} \cdot \hat{\mathbf{j}}=\hat{\mathbf{j}} \cdot \hat{\mathbf{i}}=0, \hat{\mathbf{j}} \cdot \hat{\mathbf{k}}=\hat{\mathbf{k}} \cdot \hat{\mathbf{j}}=0 ; \hat{\mathbf{k}} \cdot \hat{\mathbf{i}}=\hat{\mathbf{i}} \cdot \hat{\mathbf{k}}=0$

$\\\text {If } \theta=0, \text { then } \vec{\mathbf a} \cdot \vec{\mathbf b}=|\vec{a}||\vec{b}|\\\text {In particular, } \vec{\mathbf{a}} \cdot \vec{\mathbf{a}}=|\vec{\mathbf{a}}|^{2}\\\text {As } \hat{\mathbf{i}}, \hat{\mathbf{j}} \text { and } \hat{\mathbf{k}} \text { are unit vectors along the coordinate }\text { axes, therefore }\\\mathrm{\;\;\;\;\;\;}\hat{\mathbf{i}} \cdot \hat{\mathbf{i}}=|\hat{\mathbf{i}}|^{2}=1, \hat{\mathbf{j}} \cdot \hat{\mathbf{j}}=|\hat{\mathbf{j}}|^{2}=1 \text { and }\hat{\mathbf{k}} \cdot \hat{\mathbf{k}}=|\hat{\mathbf{k}}|^{2}=1$

Properties of Dot (Scalar) Product

$\\1.\;\;\;\;\vec{\mathbf a} \cdot \vec{\mathbf b}=\vec{\mathbf b} \cdot \vec{\mathbf a}\;\;\;\;\;\;\;\;\;\;(\text { commutative })\\2.\;\;\;\;\vec{\mathbf a}\cdot \left ( \vec{\mathbf b}+\vec{\mathbf c} \right )=\vec{\mathbf a} \cdot \vec{\mathbf c}+\vec{\mathbf a} \cdot \vec{\mathbf c}\;\;\;\;\;\;\;\;\;\;(\text { distributive })\\3.\;\;\;\;(m \vec {\mathbf a}) \cdot \vec{\mathbf{b}}=m(\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})=\vec{\mathbf{a}} \cdot(m \vec{\mathbf{b}});\text{where }\ m \text { is a scalar and } \vec a,\;\vec b \text{ are any two vectors } \\4.\;\;\;\;(l\vec{\mathbf a}) \cdot(m \vec{\mathbf b})=l m(\vec{\mathbf a} \cdot \vec{\mathbf b}) ; \text { where } l \text { and } m \text { are scalars }$

$\\\text {For any two vectors } \vec {\mathbf a} \text{ and }\vec { \mathbf b}, \;\;\text{we have }\;\\\mathrm{\;\;(i)}\;|\vec{a} \pm \vec{b}|^{2}=|\vec{a} \pm \vec{b}| \cdot|\vec{a} \pm \vec{b}|\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=}\;|\vec{a}|^{2}+|\vec{b}|^{2} \pm 2 \vec{a} \cdot \vec{b}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=}\;|\vec{a}|^{2}+|\vec{b}|^{2} \pm 2|\vec{a}||\vec{b}| \cos \theta\\\\\mathrm{\;\;(ii)}\;|\vec{a}+\vec{b}| \cdot {|\overrightarrow a}-\vec{b}|=|\vec{a}|^{2}-|\vec{b}|^{2}\\\\\mathrm{\;\;(iii)}\;\left | \vec{ a}+ \vec{ b} \right |=\left | \vec a \right |+\left | \vec b \right |\;\;\Rightarrow \;\;\vec a\,\,and\,\, \vec b\,\,are\,\,like\,\,vectors\\\\\mathrm{\;\;(iv)}\;\left | \vec{ a}+ \vec{ b} \right |=\left | \vec a -\vec b \right |\;\;\Rightarrow \;\;\vec a\perp \vec b$

Dot (Scalar) Product in Terms of Components

$\\ \text {If } \vec {\mathbf a}=a_{1} \hat{\mathbf{i}}+a_{2} \hat{\mathbf{j}}+a_{3} \hat{\mathbf{k}} \;\text { and }\; \vec {\mathbf{b}}=b_{1} \hat{\mathbf{i}}+b_{2} \hat{\mathbf{j}}+b_{3} \hat{\mathbf{k}}, \text { then } \\ {\mathbf{a} \cdot \mathbf{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}$

Proof :

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

Angle between two vectors

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{\mathbf a} \cdot \vec{\mathbf b}=|\vec{\mathbf a}||\vec{\mathbf b}| \cos \theta\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;}\cos \theta=\frac{\vec{\mathbf a} \cdot \vec{\mathbf b}}{|\vec{\mathbf a}||\vec{\mathbf b}| }\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \theta=\cos^{-1}\left (\frac{\vec{\mathbf a} \cdot \vec{\mathbf b}}{|\vec{\mathbf a}||\vec{\mathbf b}| } \right )\\\\\text{If }\;\;\vec {\mathbf a}=a_{1} \hat{\mathbf{i}}+a_{2} \hat{\mathbf{j}}+a_{3} \hat{\mathbf{k}}\;\;\text{and }\;\;\vec{\mathbf b}=b_{1} \hat{\mathbf{i}}+b_{2} \hat{\mathbf{j}}+b_{3} \hat{\mathbf{k}}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \theta=\cos ^{-1}\left(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\right)$

Geometrical Interpretation of Scalar Product

Let $\vec{\mathbf a}$ and $\vec{\mathbf b}$ be two vectors represented by OA and OB, respectively.

Draw BL ⊥ OA and AM ⊥ OB.

From triangles OBL and OAM we have OL= OB cos θ and OM = OA cos θ.

Here OL and OM are known as projections of $\vec{\mathbf b}$ on  $\vec{\mathbf a}$ and $\vec{\mathbf a}$ on $\vec{\mathbf b}$ respectively.

$\\\mathrm{Now,\;\;\;\;\;\;\;\;\;\;\;}\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=|\vec{a}|(O B \cos \theta)} \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=|\vec{a}|(O L)} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=(\text { magnitude of } \vec{a})(\text { projection of } \vec{b} \text { on } \vec{a})}\\\\\mathrm{Again,\;\;\;\;\;\;\;\;\;}\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=|\vec{b}|(|\vec{a}| \cos \theta)\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=|\vec{b}|(O A \cos \theta)} \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=|\vec{b}|(O M)} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=(\text { magnitude of } \vec{b})(\text { projection of } \vec{a} \text { on } \vec{b})}$

Thus. geometrically interpreted, the scalar product of two vectors is the product of modulus of either vectors and the projection of the other in its direction.

Thus,

$\\\text { Projection of } \vec{\mathbf a} \text { on } \vec{\mathbf b}=\frac{\vec{\mathbf a}\cdot \vec{\mathbf b}}{|\vec{\mathbf b}|}=\vec{\mathbf a} \cdot \frac{\vec{\mathbf b}}{|\vec{\mathbf b}|}=\vec{\mathbf a} \cdot \mathbf{\hat{b}}\\\\\text { Projection of } \vec{\mathbf b} \text { on } \vec{\mathbf a}=\frac{\vec{\mathbf a}\cdot \vec{\mathbf b}}{|\vec{\mathbf a}|}=\vec{\mathbf b} \cdot \frac{\vec{\mathbf a}}{|\vec{\mathbf a}|}=\vec{\mathbf b} \cdot \mathbf{\hat{a}}$

## Study it with Videos

Dot (Scalar) Product of Two Vectors
Dot (Scalar) Product in Terms of Components

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## Books

### Reference Books

#### Dot (Scalar) Product of Two Vectors

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

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