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

Concept:

\(\rm \vec a\times (\vec b\times \vec c)=(\vec a.\vec c)\vec b-(\vec a.\vec b)\vec c\)

Explanation:

\(\rm \vec a\times (2\vec b\times \vec c)=\vec b\)

⇒ \(\rm (\vec a.\vec c)2\vec b-(2\vec a.\vec b)\vec c=\vec b\)

Comparing both sides

\(\rm 2(\vec a.\vec c)=1; \vec a.\vec b=0\)

So, \(\rm \vec a\) and \(\rm \vec b\) are perpendicular.

i.e., angle between \(\rm \vec a\) and \(\rm \vec b\) is 90°

Option (1) is true.

Explanation:

Given:

\(\rm \vec a=̂ i+̂ j\ and \ \vec b=̂ j+̂ k,\)

Also, \(\vec a, \vec b, \vec c\) has same length

⇒ \(|\vec a| = |\vec b| = |\vec c|\) = √2 

Let θ be the angle between the vectors.

⇒ Cosθ = \(\frac{\vec a. \vec b}{|\vec a||\vec b|} = \frac{0+1+0}{\sqrt2} =\frac{1}{\sqrt2}\)

(I) Let \(\vec c =\hat i +\hat k\)

\(|\vec c| = \sqrt2\)

Cosθ = \(\frac{\vec a. \vec c}{|\vec a||\vec c|} = \frac{0+1+0}{\sqrt2} =\frac{1}{\sqrt2}\)

All the conditions are satisfied, so it can be vector \(\vec c\)

(II) Let \(\vec c =\rm \frac{-\hat i+4\hat j-\hat k}{3}\)

⇒ \(|\vec c| = \frac{1}{3}\sqrt18 = \sqrt2\)

Cosθ = \(\frac{\vec a. \vec c}{|\vec a||\vec c|} \)

= \(\frac{\frac{-1}{3}\frac{4}{3}}{\sqrt2 \sqrt2} = \frac{1}{2}\)

All the conditions are satisfied, so it can be vector \(\vec c\)

∴ Option (c) is correct.

Explanation:

Let P = \(\rm |\vec a\times \vec b|+√3|\vec a.\vec b|\)

= \(\rm |\vec a|.|\vec b||sinθ |+ √3|\vec a||\vec b||cosθ|\)

= \(\rm |\vec a||\vec b|[|sinθ |+ √3|cosθ|]\)

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

Now, for its maxima,

\(\frac{d}{dθ } (sinθ +√3cosθ) =0\)

⇒ Cosθ -√3 sinθ =0

⇒ \(tanθ =\frac{1}{\sqrt3}\)

⇒ 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 \(\rm \vec{a}\) and \(\rm \vec{b}\)then the angle between \(\rm \vec{a}\) and \(\rm \vec{b}\) is given by \(\cos \rm \theta = \rm \frac{\vec{a}\cdot\vec{b}}{|a||b|}\)

Calculation:

Given, angle between the vectors î - mĵ and ĵ + k̂ is \(\rm \frac x 3\)

\(⇒ \cos \rm \frac{\pi}{3} = \rm \frac{(\hat{i}-m\hat{j})\cdot(\hat{j}+\hat{k})}{|\hat{i}-m\hat{j}||\hat{j}+\hat{k}|}\)

\(⇒ \frac 1 2 = \rm \frac {-m}{\sqrt{1+m^2}\times \sqrt 2}\)

\(⇒ \rm \sqrt{1+m^2} = -\sqrt2\: m\)

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 = \(1\over 2\)

∴ θ = 60° 

Concept:

For any two vectors, \(\vec a\) and \(\vec b\) we have \(\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}}\)

Calculation:

Here, we have two vectors \(\vec a = \;\hat i - \;2\hat j + \;3\hat k\) and \(\vec b = \;3\hat i - \;2\hat j + \;\hat k\) .

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

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

By, substituting the values of \(|\vec a|\), \(|\vec b|\) and \(\vec a \cdot \vec b\) in \(\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}}\), 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 - \(\rm \overrightarrow{a} .\overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| cos\ θ \)

Vector Product of Two Vectors - \(\rm \overrightarrow{a} \times \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| sin\ θ \: \widehat{n} \)

\(\: \widehat{n}\) is the unit vector perpendicular vector, θ being the angle between  \(\rm \overrightarrow{a} \) and \(\rm \overrightarrow{b} \)

Calculation:

Given

\(\rm |\overrightarrow{a} .\overrightarrow{b}| = |\overrightarrow{a} \times \overrightarrow{b}|\)

⇒ \(\rm |\overrightarrow{a}| |\overrightarrow{b}| \cos θ = |\overrightarrow{a}| |\overrightarrow{b}| \sin θ \)

⇒ cos θ = sin θ

⇒ tan θ = 1

⇒ θ = \(\rm \frac{\pi}{4}\)

The angle between  \(\rm \overrightarrow{a} \) and \(\rm \overrightarrow{b} \) is \(\rm \frac{\pi}{4}\)

Concept:

\(\rm \vec a.\vec b=|\vec a||\vec b|\cos\theta \)

Calculation:

Here, \(\vec a + \vec b + \vec c = 0\)

\(\Rightarrow \vec b + \vec c = -\vec a\)

Taking magnitude and squaring both sides,

\(\rm \Rightarrow | \vec b + \vec c|^2 = |-\vec a|^2\)

\(\rm \Rightarrow |\vec b|^2+|\vec c|^2+2\vec b.\vec c =100\)

\(\rm \Rightarrow 2 |\vec b|.|\vec c|cos θ =100-(16 + 36)\)

\(\\ \rm \Rightarrow cosθ =\frac{48}{2\;\times\; 4 \;\times\; 6}\)

\(\Rightarrow θ =cos^{-1}(1)\)

θ = 0°

Hence, option (1) is correct. 

Concept:

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

Given: \(\vec a = 3\hat i - 2\hat j + \;\hat k\;and\;\vec b = \;\hat i - 2\hat j - 3\hat k\)

As we know that, \(\vec a \cdot \;\vec b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\)

\(\vec a \cdot \;\vec b = 3 + 4 - 3 = 4\)

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

Concept:

If θ is the angle between vectors \(\vec a\) and \(\vec b\) then 

\(\rm cosθ = \frac{⃗{a}.⃗{b}}{|⃗{a}|.|⃗{b}|}\)

Calculation:

Let the angle between vectors \(\vec a\) and \(\vec b\) be θ

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

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

On doing (i) × 2 - (ii), we get 

\(⇒ 5\vec b = 5\hat i \)

\(⇒ \vec b = \hat i \)

On putting the value of the vector \( \vec b \) in equation (i), we get

⇒ \(\rm ⃗{a} = -\hat{j}\)

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

According to the concept used

\(\rm cosθ = \frac{⃗{a}.⃗{b}}{|⃗{a}|.|⃗{b}|}\)

⇒ cosθ = 0

⇒ θ = cos^-10

⇒ \(\rm θ = \frac{\pi}{2}\)

∴ The angle between \(\rm⃗{a}\) and \(\rm⃗{b}\) is \(\rm \frac{\pi}{2}\).

Concept:

Dot Product of two vectors \(\rm \vec A\) and \(\rm \vec B\) is defined as \(\rm \vec A.\vec B=\rm \left|\vec A\right|\rm \left|\vec B\right|\cos θ\), where \(\rm \left|\vec A\right|\) is the magnitude of vector \(\rm \vec A\).

\(\rm \vec A.\vec A=\rm \left|\vec A\right|^2\).

Calculation:

We are given that "sum of two vectors \(\rm \vec a\) and \(\rm \vec b\) is a vector \(\rm \vec c\)".

⇒ \(\rm \vec b +\vec c=\vec a\)

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

⇒ \(\rm \left(\vec b +\vec c\right).\left(\vec b +\vec c\right)=\left(\vec a\right).\left(\vec a\right)\)

⇒ \(\rm \left|\vec b\right|^2+\left|\vec c\right|^2+2\vec b.\vec c=\left|\vec a\right|^2\)

Since \(\rm \left|\vec a \right|=\left|\vec b \right|=\left|\vec c \right|=4\), we get:

⇒ \(\rm 4^2+4^2+2\vec b.\vec c=4^2\)

⇒ \(\rm 16+16+2\vec b.\vec c=16\)

⇒ \(\rm 2\vec b.\vec c=-16\)

⇒ \(\rm \vec b.\vec c=-8\)

⇒ |b|.|c|.cosθ = \(-8\)

⇒ 4.4.cosθ = \(-8\)

⇒ \(cosθ = -\frac {1}{2}\)

⇒ θ = 120° 

Concept:

For any two vectors, \(\vec a\) and \(\vec b\) we have \(\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}}\)

Calculation:

Here, we have two vectors \(\vec a = \;7\hat i + \;\hat j\) and \(\vec b = \;5\hat i + \;5\hat j\) 

⇒ \(|\vec a| = \sqrt {{7^2} + {1^2}} = 5\sqrt {2}\) and \(|\vec b| = \sqrt {{5^2} + {5^2}} = 5\sqrt {2}\)

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

By, substituting the values of \(|\vec a|\), \(|\vec b|\) and \(\vec a \cdot \vec b\) in \(\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}}\), 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.

Given:

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

Concept:

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

Formula:

The angle between two vectors \(\vec{P}\) and \(\vec{Q}\) is given by : 

\(\theta = cos^{-1}(\frac{\vec{P}.\vec{Q}}{|\vec{P}||\vec{Q}|})\)

Solution:

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

= 2î + ĵ + 3k̂ 

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

∴ \(\theta = cos^{-1}(\frac{\vec{BA}.\vec{BC}}{BA.BC})\)

\(\Rightarrow \theta = cos^{-1}(\frac{9}{\sqrt 14 \sqrt 6})\))

\(\Rightarrow \theta ={{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt 84}})\)

Concept:

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

Calculation:

Given:  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors such that \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular.

As  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors

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

And  \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular

⇒ (\(\vec{\text{a}}+2 \vec{\text{b}}\)).(\(5\vec{\text{a}}−4\vec{\text{b}}\)) = 0

⇒ \(5\vec{\text{a}}.\vec{\text{a}}-4 \vec{\text{a}}.\vec{\text{b}} + 10\vec{\text{b}}.\vec{\text{a}}-8\vec{\text{b}}.\vec{\text{b}}\) = 0

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

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

⇒ \(\vec{\text{a}}.\vec{\text{b}}= {1 \over 2}\) __(iii)

Now angle between  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, 

\(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\)

⇒ \(\cos θ = {{1 \over 2} \over 1.1} = {1 \over 2}\) (from (ii) and (iii))

⇒ θ = \(\frac{\pi}{3}\)

∴ The correct option is (3).