Angle between Vectors MCQ Quiz - Objective Question with Answer for Angle between Vectors - Download Free PDF

Concept:

$\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}$

Explanation:

$\overrightarrow{\mathrm{a}}\times (2\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}$

⇒ $(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}})2\overrightarrow{\mathrm{b}}-(2\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$

Comparing both sides

$2(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}})=1;\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=0$

So, $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are perpendicular.

i.e., angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is 90°

Option (1) is true.

Explanation:

Given:

$\overrightarrow{\mathrm{a}}=̂\mathrm{i}+̂\mathrm{j}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}=̂\mathrm{j}+̂\mathrm{k},$

Also, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ has same length

⇒ $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|$ = √2 

Let θ be the angle between the vectors.

⇒ Cosθ = $\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{0+1+0}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

(I) Let $\overrightarrow{c}=\hat{i}+\hat{k}$

$|\overrightarrow{c}|=\sqrt{2}$

Cosθ = $\frac{\overrightarrow{a}.\overrightarrow{c}}{|\overrightarrow{a}||\overrightarrow{c}|}=\frac{0+1+0}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

All the conditions are satisfied, so it can be vector $\overrightarrow{c}$

(II) Let $\overrightarrow{c}=\frac{-\hat{\mathrm{i}}+4\hat{\mathrm{j}}-\hat{\mathrm{k}}}{3}$

⇒ $|\overrightarrow{c}|=\frac{1}{3}\sqrt{1}8=\sqrt{2}$

Cosθ = $\frac{\overrightarrow{a}.\overrightarrow{c}}{|\overrightarrow{a}||\overrightarrow{c}|}$

= $\frac{\frac{-1}{3}\frac{4}{3}}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

All the conditions are satisfied, so it can be vector $\overrightarrow{c}$

∴ Option (c) is correct.

Explanation:

Let P = $|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|+\surd 3|\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}|$

= $|\overrightarrow{\mathrm{a}}|.|\overrightarrow{\mathrm{b}}||\mathrm{s}\mathrm{i}\mathrm{n}\theta |+\surd 3|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}||\mathrm{c}\mathrm{o}\mathrm{s}\theta |$

= $|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|[|\mathrm{s}\mathrm{i}\mathrm{n}\theta |+\surd 3|\mathrm{c}\mathrm{o}\mathrm{s}\theta |]$

⇒ p will be max if (sin 3 + √ 3cos θ) is maximum.

Now, for its maxima,

$\frac{d}{d\theta }(sin\theta +\surd 3cos\theta )=0$

⇒ Cosθ -√3 sinθ =0

⇒ $tan\theta =\frac{1}{\sqrt{3}}$

⇒ θ =30°

∴ Option (b) is correct

Concept:

• Given: 𝐚⃗ + 𝐛⃗ + 𝐜⃗ = 0 ⇒ 𝐜⃗ = −(𝐚⃗ + 𝐛⃗)
• 𝐚⃗ and 𝐛⃗ are unit vectors ⇒ |𝐚⃗| = |𝐛⃗| = 1
• |𝐜⃗| = 2 is given
• We are to find the angle between 𝐛⃗ and 𝐜⃗

Calculation:

From the equation: 𝐜⃗ = −(𝐚⃗ + 𝐛⃗)

Take magnitude square on both sides:

|𝐜⃗|² = |𝐚⃗ + 𝐛⃗|²

⇒ |𝐜⃗|² = 𝐚⃗ · 𝐚⃗ + 𝐛⃗ · 𝐛⃗ + 2(𝐚⃗ · 𝐛⃗)

⇒ |𝐜⃗|² = 1 + 1 + 2cosθ = 2 + 2cosθ

Given: |𝐜⃗| = 2

⇒ |𝐜⃗|² = 4

⇒ 2 + 2cosθ = 4

⇒ cosθ = 1

⇒ θ = 0°

So 𝐚⃗ and 𝐛⃗ point in the same direction

Then 𝐜⃗ = −2𝐚⃗

⇒ opposite in direction to 𝐚⃗ and 𝐛⃗

Angle between 𝐛⃗ and 𝐜⃗ is 180°

∴ The correct answer is: (4) 180°

Concept:

Let the two vectors be $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$then the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is given by $\cos \theta =\frac{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}{|\mathrm{a}||\mathrm{b}|}$

Calculation:

Given, angle between the vectors î - mĵ and ĵ + k̂ is $\frac{\mathrm{x}}{3}$

$\Rightarrow \cos \frac{\pi }{3}=\frac{(\hat{\mathrm{i}}-\mathrm{m}\hat{\mathrm{j}})\cdot (\hat{\mathrm{j}}+\hat{\mathrm{k}})}{|\hat{\mathrm{i}}-\mathrm{m}\hat{\mathrm{j}}||\hat{\mathrm{j}}+\hat{\mathrm{k}}|}$

$\Rightarrow \frac{1}{2}=\frac{-\mathrm{m}}{\sqrt{1+{\mathrm{m}}^2}\times \sqrt{2}}$

$\Rightarrow \sqrt{1+{\mathrm{m}}^2}=-\sqrt{2}\mathrm{m}$

Squaring both sides, we get

⇒ 1 + m² = 2m²

⇒ m² = 1

∴ m = ± 1

Concept:

If θ is the angle between x and y then x.y = xycosθ 

Calculation:

GIven, a + b + c = 0

⇒ a + b = -c, squaring both sides we get

⇒  (a + b)² = (-c)²

⇒ |a + b|2 = |c|2

⇒ |a|² + |b|² +2|a||b|cosθ = |c|², where θ = angle betwee a and b

Putting the values we get,

3² + 5² + 2× 3× 5 cosθ = 7²

⇒ 9 + 25 + 30cosθ = 49

⇒ 30cosθ = 49 - 34 = 15

⇒ cosθ = $\frac{1}{2}$

∴ θ = 60° 

Concept:

For any two vectors, $\overrightarrow{a}$ and $\overrightarrow{b}$ we have $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|}$

Calculation:

Here, we have two vectors $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$ .

⇒ $|\overrightarrow{a}|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{14}$ and $|\overrightarrow{b}|=\sqrt{3^2+(-2{)}^2+(1{)}^2}=\sqrt{14}$

⇒ $\overrightarrow{a}\cdot \overrightarrow{b}=\left(\hat{i}-2\hat{j}+3\hat{k}\right)\cdot \left(3\hat{i}-2\hat{j}+\hat{k}\right)=3+4+3=10$

By, substituting the values of $|\overrightarrow{a}|$, $|\overrightarrow{b}|$ and $\overrightarrow{a}\cdot \overrightarrow{b}$ in $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|}$, we get

⇒ $\cos \theta =\frac{10}{\sqrt{14}\times \sqrt{14}}=\frac{5}{7}$

⇒ $\theta ={\cos }^{-1}\left(\frac{5}{7}\right)$

Hence, option B is the correct answer.

Concept:

Scalar Product of Two Vectors - $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\mathrm{c}\mathrm{o}\mathrm{s}\theta$

Vector Product of Two Vectors - $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\mathrm{s}\mathrm{i}\mathrm{n}\theta \hat{\mathrm{n}}$

$\hat{n}$ is the unit vector perpendicular vector, θ being the angle between  $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$

Calculation:

Given

$|\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|$

⇒ $|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\cos \theta =|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\sin \theta$

⇒ cos θ = sin θ

⇒ tan θ = 1

⇒ θ = $\frac{\pi }{4}$

The angle between  $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is $\frac{\pi }{4}$

Concept:

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\cos \theta$

Calculation:

Here, $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$

$\Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$

Taking magnitude and squaring both sides,

$\Rightarrow |\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}{|}^2=|-\overrightarrow{\mathrm{a}}{|}^2$

$\Rightarrow |\overrightarrow{\mathrm{b}}{|}^2+|\overrightarrow{\mathrm{c}}{|}^2+2\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=100$

$\Rightarrow 2|\overrightarrow{\mathrm{b}}|.|\overrightarrow{\mathrm{c}}|\mathrm{c}\mathrm{o}\mathrm{s}\theta =100-(16+36)$

$\Rightarrow \mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{48}{2\times 4\times 6}$

$\Rightarrow \theta =co{s}^{-1}(1)$

θ = 0°

Hence, option (1) is correct. 

Concept:

• If $\overrightarrow{a}and\overrightarrow{b}$ are two vectors, then the scalar product between the given vectors is given by: $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\times \cos \theta$
• If $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}and\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ then $\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$
• If $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ is a vector then the magnitude of the vector is given by $\left|\overrightarrow{a}\right|=\sqrt{a_1^2+a_2^2+a_3^2}$

Given: $\overrightarrow{a}=3\hat{i}-2\hat{j}+\hat{k}and\overrightarrow{b}=\hat{i}-2\hat{j}-3\hat{k}$

As we know that, $\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

$\overrightarrow{a}\cdot \overrightarrow{b}=3+4-3=4$

As we know that, if $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ is a vector then the magnitude of the vector is given by $\left|\overrightarrow{a}\right|=\sqrt{a_1^2+a_2^2+a_3^2}$

Concept:

If θ is the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ then 

$\mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{⃗\mathrm{a}.⃗\mathrm{b}}{|⃗\mathrm{a}|.|⃗\mathrm{b}|}$

Calculation:

Let the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be θ

$\overrightarrow{a}+3\overrightarrow{b}=3\hat{i}-\hat{j}$       ....(i)

$2\overrightarrow{a}+\overrightarrow{b}=\hat{i}-2\hat{j}$       ....(ii)

On doing (i) × 2 - (ii), we get 

$\Rightarrow 5\overrightarrow{b}=5\hat{i}$

$\Rightarrow \overrightarrow{b}=\hat{i}$

On putting the value of the vector $\overrightarrow{b}$ in equation (i), we get

⇒ $⃗\mathrm{a}=-\hat{\mathrm{j}}$

Now, $\overrightarrow{a}.\overrightarrow{b}=\hat{i}.(-\hat{j})=0$

According to the concept used

$\mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{⃗\mathrm{a}.⃗\mathrm{b}}{|⃗\mathrm{a}|.|⃗\mathrm{b}|}$

⇒ cosθ = 0

⇒ θ = cos^-10

⇒ $\theta =\frac{\pi }{2}$

∴ The angle between $⃗\mathrm{a}$ and $⃗\mathrm{b}$ is $\frac{\pi }{2}$.

Concept:

Dot Product of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is defined as $\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right|\cos \theta$, where $\left|\overrightarrow{\mathrm{A}}\right|$ is the magnitude of vector $\overrightarrow{\mathrm{A}}$.

$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{A}}={\left|\overrightarrow{\mathrm{A}}\right|}^2$.

Calculation:

We are given that "sum of two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is a vector $\overrightarrow{\mathrm{c}}$".

⇒ $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}$

Taking dot product of both sides with themselves, the magnitudes will still be equal:

⇒ $(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}).(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left(\overrightarrow{\mathrm{a}}\right).\left(\overrightarrow{\mathrm{a}}\right)$

⇒ ${\left|\overrightarrow{\mathrm{b}}\right|}^2+{\left|\overrightarrow{\mathrm{c}}\right|}^2+2\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}={\left|\overrightarrow{\mathrm{a}}\right|}^2$

Since $\left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right|=\left|\overrightarrow{\mathrm{c}}\right|=4$, we get:

⇒ $4^2+4^2+2\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=4^2$

⇒ $16+16+2\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=16$

⇒ $2\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=-16$

⇒ $\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}=-8$

⇒ |b|.|c|.cosθ = $-8$

⇒ 4.4.cosθ = $-8$

⇒ $cos\theta =-\frac{1}{2}$

⇒ θ = 120° 

Concept:

For any two vectors, $\overrightarrow{a}$ and $\overrightarrow{b}$ we have $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|}$

Calculation:

Here, we have two vectors $\overrightarrow{a}=7\hat{i}+\hat{j}$ and $\overrightarrow{b}=5\hat{i}+5\hat{j}$ 

⇒ $|\overrightarrow{a}|=\sqrt{7^2+1^2}=5\sqrt{2}$ and $|\overrightarrow{b}|=\sqrt{5^2+5^2}=5\sqrt{2}$

⇒ $\overrightarrow{a}\cdot \overrightarrow{b}=\left(7\hat{i}+\hat{j}\right)\cdot \left(5\hat{i}+5\hat{j}\right)=35+5=40$

By, substituting the values of $|\overrightarrow{a}|$, $|\overrightarrow{b}|$ and $\overrightarrow{a}\cdot \overrightarrow{b}$ in $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|}$, we get

⇒ $\cos \theta =\frac{40}{5\sqrt{2}\times 5\sqrt{2}}=\frac{4}{5}$

⇒ $\theta ={\cos }^{-1}\left(\frac{4}{5}\right)$

Hence, option B is the correct answer.

Concept:

• The angle between two vectors  $\overrightarrow{\text{a}}$ and $\overrightarrow{\text{b}}$ is given by, $\cos \theta =\frac{\overrightarrow{\text{a}}.\overrightarrow{\text{b}}}{|\overrightarrow{\text{a}}||\overrightarrow{\text{b}}|}$ 
• If $\overrightarrow{\text{a}}$ and $\overrightarrow{\text{b}}$ are perpendicular vectors, then $\overrightarrow{\text{a}}$.$\overrightarrow{\text{b}}$ = 0 
• For any two vectors $\overrightarrow{\text{a}}$ and $\overrightarrow{\text{b}}$, $\overrightarrow{\text{a}}.\overrightarrow{\text{b}}=\overrightarrow{\text{b}}.\overrightarrow{\text{a}}$ __(i)

Calculation:

Given:  $\overrightarrow{\text{a}}$ and $\overrightarrow{\text{b}}$ are two unit vectors such that $\overrightarrow{\text{a}}+2\overrightarrow{\text{b}}$ and $5\overrightarrow{\text{a}}-4\overrightarrow{\text{b}}$ are perpendicular.

As  $\overrightarrow{\text{a}}$ and $\overrightarrow{\text{b}}$ are two unit vectors

⇒  $|\overrightarrow{\text{a}}|=|\overrightarrow{\text{b}}|=1$___(ii)

And  $\overrightarrow{\text{a}}+2\overrightarrow{\text{b}}$ and $5\overrightarrow{\text{a}}-4\overrightarrow{\text{b}}$ are perpendicular

⇒ ($\overrightarrow{\text{a}}+2\overrightarrow{\text{b}}$).($5\overrightarrow{\text{a}}-4\overrightarrow{\text{b}}$) = 0

⇒ $5\overrightarrow{\text{a}}.\overrightarrow{\text{a}}-4\overrightarrow{\text{a}}.\overrightarrow{\text{b}}+10\overrightarrow{\text{b}}.\overrightarrow{\text{a}}-8\overrightarrow{\text{b}}.\overrightarrow{\text{b}}$ = 0

⇒ $5|\overrightarrow{\text{a}}{|}^2-4\overrightarrow{\text{a}}.\overrightarrow{\text{b}}+10\overrightarrow{\text{a}}.\overrightarrow{\text{b}}-8|\overrightarrow{\text{b}}{|}^2$ = 0 {from (i)}

⇒ $5(1)+6\overrightarrow{\text{a}}.\overrightarrow{\text{b}}-8(1)$ = 0 {from (ii)}

⇒ $\overrightarrow{\text{a}}.\overrightarrow{\text{b}}=\frac{1}{2}$ __(iii)

Now angle between  $\overrightarrow{\text{a}}$ and $\overrightarrow{\text{b}}$ is given by, 

$\cos \theta =\frac{\overrightarrow{\text{a}}.\overrightarrow{\text{b}}}{|\overrightarrow{\text{a}}||\overrightarrow{\text{b}}|}$

⇒ $\cos \theta =\frac{\frac{1}{2}}{1.1}=\frac{1}{2}$ (from (ii) and (iii))

⇒ θ = $\frac{\pi }{3}$

∴ The correct option is (3).

Given:

The vertices A,B,C of a triangle ABC are (1, 1, 3), (-1, 0, 0), (0, 1, 2) respectively.

Concept:

$\overrightarrow{PQ}$ = p.v.($\overrightarrow{Q}$) - p.v.($\overrightarrow{P}$)

Formula:

The angle between two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is given by : 

$\theta =co{s}^{-1}(\frac{\overrightarrow{P}.\overrightarrow{Q}}{|\overrightarrow{P}||\overrightarrow{Q}|})$

Solution:

$\overrightarrow{BA}$ = (1, 1, 3) - (-1, 0, 0)

= 2î + ĵ + 3k̂ 

$\overrightarrow{BC}$ = (0, 1, 2) - (-1, 0, 0) = î + ĵ + 2k̂ 

∴ $\theta =co{s}^{-1}(\frac{\overrightarrow{BA}.\overrightarrow{BC}}{BA.BC})$

$\Rightarrow \theta =co{s}^{-1}(\frac{9}{\sqrt{1}4\sqrt{6}})$)

$\Rightarrow \theta ={\cos }^{-1}(\frac{9}{\sqrt{8}4})$