Dot Product of Two Vectors Questions

The dot product of two vectors (also called the scalar product) is a fundamental operation in vector algebra, widely used in physics, engineering, and mathematics. It helps in finding the angle between vectors, determining projections, and checking orthogonality. In this blog, we will explore the definition, formula, properties, and most importantly, solved questions related to the dot product of two vectors, including some challenging JEE and JEE Advanced level problems.

1.0Definition of Dot Product

The dot product of two vectors a and b is defined as: $a.b=|a||b|Cos\theta$

where:

• $|a|$ = magnitude of vector a
• $|b|$ = magnitude of vector b
• $\theta$= angle between vectors a and b

2.0Dot Product in Component Form

If: $a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$, then $a.b=a_1{b}_1+a_2{b}_2+a_3{b}_3$

Key Properties

1. Commutative: $a.b=b.a$
2. Distributive over addition: $a.(b+c)=a.b+a.c$
3. Orthogonal vectors: If $a.b=0$, then a and b are perpendicular.
4. Relation with projection: $a.b=|a|\times \text{(projection of b on a )}$

3.0Dot Product of Two Vectors Solved Questions

Example 1: Find the dot product of $\mathbf{a}=2i+3j-k\text{and}\mathbf{b}=i-2j+4k.$

Solution:

$\mathbf{a}\cdot \mathbf{b}=(2)(1)+(3)(-2)+(-1)(4)$

$=2-6-4=-8$

Answer: -8

Example 2: If $\mathbf{a}=i+2j+2k\text{and}\mathbf{b}=2i+j+2k$, find the angle between them.

Solution:

$\mathbf{a}\cdot \mathbf{b}=(1)(2)+(2)(1)+(2)(2)=2+2+4=8$

$|\mathbf{a}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3$

$|\mathbf{b}|=\sqrt{2^2+1^2+2^2}=\sqrt{9}=3$

$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{8}{3\times 3}=\frac{8}{9}$

$\theta =co{s}^{-1}(\frac{8}{9})$

Answer: $\theta =co{s}^{-1}(\frac{8}{9})$

Example 3: Let $P=3i-2j+kandq=\lambda i+j+2k$. If P is perpendicular to Q, find $\lambda$.

Solution:

For perpendicular vectors: $p.q=0$

$(3)(\lambda )+(-2)(1)+(1)(2)=0$

$3\lambda -2+2=0$

$3\lambda =0\Rightarrow =0$

Answer: $\lambda =0$

Example 4: $p=3i-2j+k,q=\lambda i+j+2k$. If $p\perp q$, find $\lambda$.

Solution:

$p.q=0$

$(3)(\lambda )+(-2)(1)+(1)(2)=0$

$3\lambda -2+2=0$

$3\lambda =0\Rightarrow \lambda =0$

Answer: $\lambda =0$

Example 5: $|a|=3,|b|=5,a\cdot b=15$. Find the angle between a and b, and the length of projection of b on a.

Solution:

$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{15}{3.5}=1\Rightarrow \theta =0^o$

Projection length of b on a is $|b|cos\theta =5$

Answer: $\theta =0^o$, projection length =5.

Example 6: $\mathbf{a}=2i-j+2k,\mathbf{b}=i+2j+2k$. Find the angle between them.

Solution:

$\mathbf{a}\cdot \mathbf{b}=2(1)+(-1)(2)+2(2)=2-2+4=4.$

$|\mathbf{a}|=\sqrt{4+1+4}=3,|\mathbf{b}|=\sqrt{1+4+4}=3$

$cos\theta =\frac{4}{3.3}=\frac{4}{9}$

$\theta =co{s}^{-1}(\frac{4}{9})$

Answer: $\theta =co{s}^{-1}(\frac{4}{9})$.

Example 7: If $a,b\ne 0$ and $|a+b|=|a-b|$, show that $a.b=0$.

Solution:

Square both sides:

$|\mathbf{a}+\mathbf{b}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+2\mathbf{a}\cdot \mathbf{b}$

$|\mathbf{a}-\mathbf{b}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2\mathbf{a}\cdot \mathbf{b}$

Equality gives $+2a.b=-2a.b\Rightarrow a.b=0$

Answer: $a\perp b$

Example 8: For $\mathbf{a}=(3,-1,2)\text{and}\mathbf{b}=(1,2,2)$, find the projection of a on b and the component of a perpendicular to b.

Solution:

$\mathbf{a}\cdot \mathbf{b}=3(1)+(-1)(2)+2(2)=5,|\mathbf{b}{|}^2=1^2+2^2+2^2=9$

${\text{proj}}_{\mathbf{b}}(\mathbf{a})=\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{b}{|}^2}\mathbf{b}$

$=\frac{5}{9}<1,2,2>=⟨\frac{5}{9},\frac{10}{9},\frac{10}{9}⟩$

Perpendicular component: 

$a_{\perp }=a-pro{j}_b(a)$

$=<3,-1,2>-<\frac{5}{9},\frac{10}{9},\frac{10}{9}>$

$=<\frac{22}{9},\frac{19}{9},\frac{8}{9}>$

Answer: 

$pro{j}_b(a)=<\frac{5}{9},\frac{10}{9},\frac{10}{9}>$

$a_{\perp }==<\frac{22}{9},\frac{19}{9},\frac{8}{9}>$.

Example 9: A force $\mathbf{F}=(1,-2,2)$ moves a particle through displacement $s=(2,1,-4)m$. Find the work done.

Solution:

$W=\mathbf{F}\cdot \mathbf{s}=6.2+(-2).1+3.(-4)=12-2=-2J$.

Answer: -2 J.

Example 10: Problem: A parallelogram has adjacent sides a and b. Find the angle between its diagonals a+b and a-b.

Solution:

$(a+b).(a-b)=|a{|}^2-|b{|}^2$

Also $|\mathbf{a}+\mathbf{b}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+2\mathbf{a}\cdot \mathbf{b},|\mathbf{a}-\mathbf{b}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2\mathbf{a}\cdot \mathbf{b}.$

Thus $\cos \theta =\frac{|\mathbf{a}{|}^2-|\mathbf{b}{|}^2}{\sqrt{|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+2\mathbf{a}\cdot \mathbf{b}}\sqrt{|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2\mathbf{a}\cdot \mathbf{b}}}.$

Answer: $\theta =co{s}^{-1}(\frac{|\mathbf{a}{|}^2-|\mathbf{b}{|}^2}{\sqrt{|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+2\mathbf{a}\cdot \mathbf{b}}\sqrt{|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2\mathbf{a}\cdot \mathbf{b}}}.)$

Example 11: Find the angle between the space diagonals $\overrightarrow{OA}$ and $\overrightarrow{OB}$ of a cube of side a with O = (0, 0, 0), A = (a, a, a), B = (a, -a, a).

Solution:

$\overrightarrow{OA}=⟨a,a,a⟩,\overrightarrow{OB}=⟨a,-a,a⟩.$

Dot product = $\overrightarrow{OA}\cdot \overrightarrow{OB}=a^2-a^2+a^2=a^2.$

$|\overrightarrow{OA}|=|\overrightarrow{OB}|=\sqrt{a^2+a^2+a^2}=a\sqrt{3}.$

$\cos \theta =\frac{\overrightarrow{OA}\cdot \overrightarrow{OB}}{|\overrightarrow{OA}||\overrightarrow{OB}|}=\frac{a^2}{a\sqrt{3}\cdot a\sqrt{3}}=\frac{1}{3}.$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$

Answer: $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$

Example 12: For fixed $a\ne 0$ and fixed k>0, maximize a.x subject to $|x|=k$. Also find the maximizing x.

Solution:

By Cauchy–Schwarz, $|\mathbf{a}\cdot \mathbf{x}|\le |\mathbf{a}||\mathbf{x}|=k|\mathbf{a}|.$

Maximum value is $k|a|$ when x is in the direction of a.

So $x=k\frac{a}{|a|}$

Answer: Max $=k|a|$, attained at $x=ka/|a|$.

Example 13: Unit vectors $a,b,c$ satisfy $a.b=b.c=c.a=\frac{1}{3}$ Find the common angle between each pair.

Solution:

For unit vectors, $a.b=cos\theta$ Given $cos\theta =\frac{1}{3}$

Hence $\theta =co{s}^{-1}(\frac{1}{3})$

Answer: $\theta =co{s}^{-1}(\frac{1}{3})$ for each pair.

Example 14: $u=(\lambda ,2,-1)$ is a unit vector and $v=<1,-1,2>$. If u⋅v=0 , find $\lambda$.

Solution:

Orthogonality: 

$\lambda (1)+2(-1)+(-1)(2)=0$

$\Rightarrow \lambda -2-2=0$

$\lambda =4$

Check unit: 

$$\begin{gathered} |u{|}^2={\lambda }^2+4+1 \\ =16+5 \\ =21\ne 1 \end{gathered}$$

This violates unit length, so no λ  satisfies both unless we scale. Fix by imposing unit first:

$$\begin{gathered} |u{|}^2={\lambda }^2+5=1 \\ \Rightarrow {\lambda }^2=-4 \end{gathered}$$

impossible over reals.
Therefore no real λ makes u both unit and orthogonal to with this template.

Answer: No real solution.

Example 15: Vectors a, b satisfy ∣a∣=∣b∣=5 and ∣a+b∣=6 . Find the angle between a and b.

Solution:

$$\begin{gathered} \mid a+b{\mid }^2=\mid a{\mid }^2+\mid b{\mid }^2+2a\cdot b \\ 36=25+25+2a.b \\ \Rightarrow 2a.b=-14 \\ \Rightarrow a.b=-7 \end{gathered}$$

$$\begin{gathered} cos\theta =\frac{-7}{5.5}=\frac{7}{25} \\ \theta =co{s}^{-1}(-\frac{7}{25}) \end{gathered}$$

Answer: $\theta =co{s}^{-1}(-\frac{7}{25})$

Example 16: Check if $u_1=\frac{1}{\sqrt{2}}<1,1,0>,u_2=\frac{1}{\sqrt{6}}<1,-1,2>,u_3=\frac{1}{\sqrt{3}}<1,1,1>$ are mutually orthonormal.

Solution:

Each is unit by construction.

$u_1\cdot u_2=\frac{1}{\sqrt{12}}(1\cdot 1+1\cdot (-1)+0\cdot 2)=\frac{0}{\sqrt{12}}=0$

$u_1\cdot u_3=\frac{1}{\sqrt{6}}(1\cdot 1+1\cdot 1+0.1)=\frac{2}{\sqrt{6}}=\ne 0$

So not mutually orthonormal.

Answer: Not orthonormal since $u_1.u_3\ne 0$.

Example 17: $a=<1,2,2>$A vector $b=<t,1,-1>$ has projection length on a equal to 2. Find t.

Solution:

Projection length of b on a is $\frac{|a.b|}{|a|}=2$

$$\begin{gathered} a.b=1.t+2.1+2.(-1)=t+2-2=t. \\ |a|=\sqrt{1+4+4}=3 \end{gathered}$$

So, $\frac{|t|}{3}=2\Rightarrow |t|=6\Rightarrow t=\pm 6$

Answer: t = 6 or t = -6.

4.0Practice Questions on Dot Product of Two Vectors Questions

1. Find the dot product of $a=4i-j+2kandb=-i+3j-5k$.
2. Two vectors have magnitudes 7 and 10 with an angle of 60° between them. Find their dot product.
3. If $a.b=15,|a|=5,and|b|=4$, find the angle between a and b.
4. Vectors $u=i+j+kandv=xi+2j+3k$ are perpendicular. Find x.