Vector Algebra MCQ Quiz - Objective Question with Answer for Vector Algebra - Download Free PDF

Calculation:

Given,

The vectors \(\vec{a},\vec{b} ,(\vec{a}\times\vec{b})\) are unit vectors.

Since \(\vec a\) and  \(\vec{b}\) are unit vectors, we know:

\( |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \).

The magnitude of the cross product \(( \vec{a} \times \vec{b} )\) is given by:

\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \times 1 \times \sin \theta = \sin \theta \).

Since \( ( |\vec{a} \times \vec{b}| = 1 \), we have:

\( \sin \theta = 1 \), so \(\theta = 90^\circ \), meaning \(\vec a \) and \( \vec{b} \) are perpendicular.

The dot product \( ( \vec{a} \cdot \vec{b} )\) is:

\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \times 1 \times \cos 90^\circ = 0 \).

∴ The value of\(( \vec{a} \cdot \vec{b} ) \) is 0.

Hence, the correct answer is Option 1.

Calculation:

Given,

The position vectors of points A, B, and C are \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) respectively, and \( \vec{c} = \cos^2 \theta \, \vec{a} + \sin^2 \theta \, \vec{b} \).

The expression to evaluate is: \( (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) \).

First, substitute \( \vec{c} \) into the equation:

\( (\vec{a} \times \vec{b}) + (\vec{b} \times (\cos^2 \theta \, \vec{a} + \sin^2 \theta \, \vec{b})) + (\cos^2 \theta \, \vec{a} + \sin^2 \theta \, \vec{b}) \times \vec{a} \).

Using the distributive property of the cross product:

\( (\vec{a} \times \vec{b}) + \left[ (\vec{b} \times \cos^2 \theta \, \vec{a}) + (\vec{b} \times \sin^2 \theta \, \vec{b}) \right] + \left[ (\cos^2 \theta \, \vec{a} \times \vec{a}) + (\sin^2 \theta \, \vec{b} \times \vec{a}) \right] \).

Since \( \vec{b} \times \vec{b} = 0 \) and \( \vec{a} \times \vec{a} = 0 \), we are left with:

\( (\vec{a} \times \vec{b}) + \cos^2 \theta \, (\vec{b} \times \vec{a}) + \sin^2 \theta \, (- \vec{a} \times \vec{b}) \).

Substitute \( \vec{b} \times \vec{a} = - (\vec{a} \times \vec{b}) \) into the expression:

\( (\vec{a} \times \vec{b}) + \cos^2 \theta \, (- \vec{a} \times \vec{b}) + \sin^2 \theta \, (- \vec{a} \times \vec{b}) \).

Factor out \( \vec{a} \times \vec{b} \):

\( \vec{a} \times \vec{b} \left[ 1 - \cos^2 \theta - \sin^2 \theta \right] \).

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), the expression becomes:

\( \vec{a} \times \vec{b} [1 - 1] = 0 \).

∴ The final result is \( \vec{0} \).

Hence, the correct answer is option 1. 

Calculation:

Given,

\( 3\vec{a} - 4\vec{b} + \vec{c} = 0 \)

\( \vec{c} = 4\vec{b} - 3\vec{a} \)

The vector\(\overrightarrow{AB} \) is:

\( \overrightarrow{AB} = \vec{b} - \vec{a} \)

The vector \(\overrightarrow{BC} \) is:

\( \overrightarrow{BC} = \vec{c} - \vec{b} \)

Substituting \(\vec{c} = 4\vec{b} - 3\vec{a} \):

\( \overrightarrow{BC} = (4\vec{b} - 3\vec{a}) - \vec{b} \)

\( \overrightarrow{BC} = 3\vec{b} - 3\vec{a} \)

Step 4: Now, \(\overrightarrow{BC} = 3(\vec{b} - \vec{a}) \), which gives:

\( AB : BC = 1 : 3 \)

∴ The correct ratio is AB : BC = 1 : 3 , 

Hence, the correct answer is Option 2. 

Calculation:

Given,

The vector \( \vec{d} = (\vec{a} \times \vec{b}) \times \vec{c} \)

Statement I: \( \vec{d} \) is coplanar with \( \vec{a} \) and \( \vec{b} \).

We use the vector triple product identity: \( (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} \).

This shows that \( \vec{d} \) is a linear combination of \( \vec{a} \) and \( \vec{b} \), hence \( \vec{d} \) is coplanar with \( \vec{a} \) and \( \vec{b} \).

Therefore, Statement I is correct.

Statement II: \( \vec{d} \) is perpendicular to \( \vec{c} \).

To check this, compute the dot product \( \vec{d} \cdot \vec{c} \). Using the vector triple product identity, we find:

\( \vec{d} \cdot \vec{c} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c}) = 0 \),

which means \( \vec{d} \) is perpendicular to \( \vec{c} \).

Therefore, Statement II is correct.

∴ Both Statement I and Statement II are correct.

Hence, the correct answer is option  3. 

Calculation:

Given,

\( \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1 \)

Using the identity \( \cos^2(x) = 1 - \sin^2(x) \), we substitute:

\( (1 - \sin^2(\alpha)) + (1 - \sin^2(\beta)) + (1 - \sin^2(\gamma)) = 1 \)

Simplifying the equation:

\( 3 - (\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma)) = 1 \)

Rearrange to isolate the sine terms:

\( \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2 \)

Now, calculate the dot product:

\( \vec{a} \cdot \vec{b} = \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2 \)

∴ The value of \( \vec{a} \cdot \vec{b} \)is 2.

Hence, the correct answer is Option 4.

Concept:

Conditions of collinear vector:

• Three points with position vectors \(\vec a,\;\vec b\;and\;\vec c\) are collinear if and only if the vectors \(\left( {\vec a - \vec b} \right)\) and \(\left( {\vec a\; - \vec c} \right)\) are parallel. ⇔ \(\left( {\vec a - \vec b} \right) = λ \left( {\vec a\; - \vec c} \right)\)
• If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Solution:

We know that, If the points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) be collinear then \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Given  \(\widehat i + 2\widehat j + 3\widehat k\), \(λ \widehat i + 4\widehat j + 7\widehat k\), \(- 3\widehat i - 2\widehat j - 5\widehat k\) are collinear

∴ \(\left| {\begin{array}{*{20}{c}} { 1}&{ 2}&3\\ λ&4&7\\ -3&-2&-5 \end{array}} \right| = 0\)

⇒ 1 (-20 + 14) – (2) (-5λ + 21) + 3 (-2λ + 12) = 0

⇒ -6 + 10λ – 42 - 6λ + 36  = 0

⇒ 4λ = 12

∴ λ = 3

Concept:

Let \(\rm {\rm{\vec a}} = {\rm{x\;\vec i}} + {\rm{y\;\vec j}} + {\rm{z\;\vec k}}\) then magnitude of the vector of a = \(\left| {{\rm{\vec a}}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {\rm{\;}}{{\rm{y}}^2} + {{\rm{z}}^2}} \)

Calculation:

Let \(\rm \vec{a}\) =  p(2î - ĵ + 2k̂)

Given, \(\left| {{\rm{\vec a}}} \right| = 3\)

⇒ \(\rm \sqrt{4p^2 + p^2+4p^2} = 3\)

⇒ \(\rm \sqrt{9p^2} = 3\)

⇒ 3p = 3

∴ p = 1

Concept:

Dot product of two vectors is defined as:

\({\rm{\vec A}}{\rm{.\vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times cos}}\;{\rm{\theta }}\)

Cross/Vector product of two vectors is defined as:

\({\rm{\vec A \times \vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times sin}}\;{\rm{\theta }} \times \rm \hat{n}\)

where θ is the angle between \({\rm{\vec A}}\;{\rm{and}}\;{\rm{\vec B}}\)

Calculation:

To Find: Value of \(\rm \vec{a} \times \vec{a}\)

Here angle between them is 0°

\({\rm{\vec a \times \vec a = }}\left| {\rm{a}} \right|{\rm{ \times }}\left| {\rm{a}} \right|{\rm{ \times sin}}\;{\rm{0 }} \times \rm \hat{n}=0\)

Concept:

If \(\rm \vec A = x \hat i-y\hat j +z\hat k\), then \(\rm |\vec A| = \sqrt {x^2 +y^2+z^2}\)

Calculation:

Given A = \(\rm 5 \hat i-2\hat j +4\hat k\) and B = \(\rm \hat i+3\hat j -7\hat k\)

\(\rm \vec{AB} = \vec B - \vec A\)

\(\rm \vec{AB}\) = \(\rm \hat i+3\hat j -7\hat k - (5 \hat i-2\hat j +4\hat k)\)

\(\rm \vec{AB}\) = \(\rm -4\hat i+5\hat j -11\hat k\)

Now \(\rm |\vec {AB}| = \sqrt{(-4)^2 +5^2+(-11)^2}\)

\(\rm |\vec {AB}| = \sqrt{16 +25+121}\)

\(\rm |\vec {AB}| = \sqrt{162}\) = 9√2

Concept:

Three or more points are collinear, if slope of any two pairs of points is same.

Here, \(\rm 5\hat i-2\hat j, 8\hat i-3\hat j, a\hat i-12\hat j \)

Let, A = (5, -2), B = (8, -3), C = (a, -12)

Now, slope of AB = Slope of BC = Slope of AC ....(∵ points are collinear)

 \(\rm \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{a-8}\\ ⇒ \frac{-1}{3}=\frac{-9}{a-8}\)

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.

Concept:

For two vectors \(\vec m \ and \ \vec n \) to be collinear,​ \(\vec m\; = \;λ \vec n\) where λ is a scalar.

Calculation:

Given that, the vectors \(4\hat i + \hat j - 3\hat k\) & \(p\hat i + q\hat j - 2\hat k\) are collinear.

Since two vectors \(\vec m \ and \ \vec n \) are collinear then \(\vec m\; = \;λ \vec n\) where λ is a scalar.

⇒ \(4\hat i + \hat j - 3\hat k\;\ = λ × (\;p\hat i + q\hat j - 2\hat k)\)

⇒ \(4\hat i + 1\hat j - 3\hat k\;\ = λ p \hat i + λq \hat j - 2λ \hat k\)

⇒ λp = 4,  λq = 1 and -2λ = -3

⇒  λ = 3/2 

So, by substituting λ = 3/2 in  λp = 4 and λq = 1, we get

⇒ (3/2)p = 4 and (3/2)q = 1

⇒ p = 8/3 and q  = 2/3

∴  \(\frac{8}{3}, \frac{2}{3}\)is the correct answer.

Concept:

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 \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos \theta\)

Calculation:

Given: \(\vec a = 2\hat i - 6\hat j - 3\hat k\) and \(\vec b = 4\hat i + 3\hat j - \hat k\)

\(\left| {\vec a} \right| = 7,\;\left| {\vec b} \right| = \sqrt {26} \;and\;\vec a \cdot \;\vec b = - 7\)

\(\Rightarrow \;\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}} = \frac{{ - \;7}}{{7 \times \sqrt {26} }} = - \frac{1}{{\sqrt {26} }}\)

\( \Rightarrow \;{\sin ^2}\theta = 1 - {\cos ^2}\theta = 1 - \frac{1}{{26}} = \frac{{25}}{{26}}\)

Concept:

Let the angle between \(\vec a\) and \(\vec b\)is \(\rm \theta\)

\(\rm \vec a.\vec b = 2ab cos\;\theta\)

Calculations:

consider, the angle between \(\vec a\) and \(\vec b\)is \(\rm \theta\)

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

⇒\(\vec a + \vec b = - \vec c \)

⇒\(\rm |\vec a + \vec b| = |- \vec c |\)

Squaring on both side, we get

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

⇒\(\rm |\vec a|^2 +2\;\vec a.\vec b+ |\vec b|^2 = |- \vec c |^2\)

⇒\(\rm |\vec a|^2 +|\vec b|^2+2\;ab\cos\;\theta = |- \vec c |^2\)

⇒\(\rm (3)|^2 +(5)^2+2\;(3)(5)\cos\;\theta = (7)^2\)

⇒\(\rm 30\cos\;\theta = 15\)

⇒\(\rm \cos\;\theta = \dfrac 12\)

⇒ \(\rm \theta\) = π / 3

Hence, If \(\vec a + \vec b + \vec c = \vec 0,\;|\vec a| = 3,\;|\vec b| = 5\) and \(|\vec c| = 7\), then the angle between \(\vec a\) and \(\vec b\)is π / 3

Concept:

If \({\rm{\vec a\;and\;\vec b}}\) are two vectors parallel to each other then \({\rm\vec{a} = λ \vec{b}}\) or \(\rm \vec{a} × \vec{b} =0\)

Calculation:

Given:

 \(\rm \vec{i} - a\vec{j} + 5\vec{k}\) and \(\rm 3\vec{i} - 6\vec{j} + b\vec{k}\) are parallel vectors,

Therefore, \(\rm \vec{i} - a\vec{j} + 5\vec{k}= λ (\rm 3\vec{i} - 6\vec{j} + b\vec{k})\)

Equating the coefficient of \(\rm \vec{i},\vec{j} \;and\; \vec{k}\)

⇒ 1 = 3λ, ∴ λ = 1/3            

⇒ -a = -6λ 

⇒ 5 = bλ                 .... (1)

Put the value of λ in equation (1), we get

5 = b × (1/3)

So, b = 15

Calculation:

 \(\rm \vec a =\hat i +\hat j +\hat k,\; \vec b =\hat i -\hat j + \hat k\) and c = î - ĵ - k̂

Given:  vector \(\rm \vec v\) in the plane of \(\rm \vec a\) and \(\rm \vec b\) 

Therefore, \(\rm \vec v = \vec a + λ \vec b\)

⇒ \(\rm \vec v =(\hat i +\hat j +\hat k ) \; + λ (\hat i -\hat j + \hat k)\)

= (1 + λ)î + (1 - λ)ĵ  + (1 + λ)k̂                .... (1)

Projection of \(\rm \vec v\) on \(\rm \frac {\vec c} {|\vec c|}=\frac 1 {\sqrt 3}\)

⇒ \(\rm \vec v=\rm \frac {\vec c} {|\vec c|}=\frac 1 {\sqrt 3}\) 

⇒ \(\frac {(1 + λ) - (1 - λ) - (1 + λ)}{\sqrt3} = \frac {1}{\sqrt 3}\)

⇒ -(1 - λ) = 1

∴ λ = 2

Now, put the value of λ in equation (1), we get 

\(\rm \vec v\) = 3î - ĵ + 3k̂