Properties of Vectors MCQ Quiz - Objective Question with Answer for Properties of Vectors - Download Free PDF

Concept:

Area of parallelogram determined by the the vectors \(\vec a\) and \(\vec b\) is |\(\vec a\) × \(\vec b\)|.

Explanation:

Given

\(\vec a\) = î + 2ĵ +3k̂ and \(\vec b\) = 3î - 2ĵ + k̂

So, 

\(\vec a\) × \(\vec b\) = \(\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\3&-2&1\end{vmatrix}\)

    = î(2 + 6) + ĵ(9 - 1) + k̂(-2 - 6) = 8î + 8ĵ - 8k̂

Hence area of the parallelogram 

= |\(\vec a\) × \(\vec b\)| = \(\sqrt{8^2+8^2+(-8)^2}\) = \(\sqrt{64+64+64}\)  = 8√3

Option (1) is true.

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 \(\vec b = 4\hat i + 3\hat j - \hat k\)

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

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

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

Calculation

From triangular law of vector addition, we get \(OP+PS = OS\)

\(\therefore \vec P+b|\vec R|\dfrac{\vec R}{|\vec R|}=\vec{S}\)

⇒ \(\vec P+b{\vec R}=\vec{S}\)

But \(\vec{R} = \vec{Q} - \vec{P}\) (Given)

⇒ \(\vec P+b(\vec Q-\vec P)=\vec{S}\)

⇒ \(\vec{S} =(1-b) \vec P+b \vec Q\)

Hence option 3 is correct

Calculation

\( [\vec{a} \ \vec{b} \ \vec{y}] = (\vec{a} \times \vec{b}) \cdot \vec{y} = \vec{a} \begin{vmatrix} i & j & k \\ 1 & 1 & -1 \\ 1 & 0 & 1 \end{vmatrix} \)

\( \vec{a} = (i + 2j - k) \) is maximum → angle between \( \vec{a} \) & \( i + 2j - k \) will be 0

⇒  \( [\vec{a} \ \vec{b} \ \vec{y}] = |\vec{a}||i + 2j - k| = \sqrt{6} \)

Hence option 4 is correct

Calculation: 

We know that if two vectors \(\rm\overrightarrow{a}\&\overrightarrow{b}\) are orthogonal then, \(\rm\overrightarrow{a}⋅\overrightarrow{b}\) = 0

since, (î − xĵ − 2k̂) & (2î + ĵ + yk̂) are orthogonal then, (î − xĵ − 2k̂)⋅(2î + ĵ + yk̂) = 0

⇒ 2 − x − 2y = 0

⇒ x + 2y = 2 represents a straight line.

Hence, locus of point (x, y) is a straight line.

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:

• \({\left| {\vec a - \vec b} \right|^2} + \;{\left| {\vec a + \vec b} \right|^2} = \;2\; \times \;\left( {{{\left| {\vec a} \right|}^2} + \;{{\left| {\vec b} \right|}^2}} \right)\)

Calculation:

Given:  \(|\vec a|\) = 3, \(\left| {\vec b} \right| = 4\) and \(\left| {\vec a - \vec b} \right| = 5,\)

We know that,

\({\left| {\vec a - \vec b} \right|^2} + \;{\left| {\vec a + \vec b} \right|^2} = \;2\; \times \;\left( {{{\left| {\vec a} \right|}^2} + \;{{\left| {\vec b} \right|}^2}} \right)\)

\(\Rightarrow {5^2} + \;{\left| {\vec a + \vec b} \right|^2} = \;2\; \times \;\left( {{3^2} + \;{4^2}} \right)\)

\(\Rightarrow {\left| {\vec a + \vec b} \right|^2} = \;50 - 25 = 25\)

∴ \(\left| {\vec a + \vec b} \right| = 5\)

Concept:

• \(\vec a\) and \(\vec b\) are two vectors parallel to each other ⇔  \(\vec a \times \vec b = 0\)
• Cross product of parallel vectors are zero ⇔  \(\vec a \times \vec a = 0,{\rm{\;}}\vec b \times \vec b = 0{\rm{\;and\;}}\vec c \times \vec c = 0\)
• A cross or vector product is not commutative ⇔ \(\vec a \times \vec b = - {\rm{\;}}\vec b \times \vec a\)

Calculation:

We have to find the value of \(\left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right)\)

\(\Rightarrow \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) = {\rm{\;\vec a\;}} \times {\rm{\;\vec a}} + {\rm{\;\vec a\;}} \times {\rm{\;\vec b}} - {\rm{\;\vec b\;}} \times {\rm{\;\vec a}} - {\rm{\;\vec b\;}} \times {\rm{\;\vec b}}\)

We know that \({\rm{\vec a\;}} \times {\rm{\;\vec b}} = - {\rm{\;\vec b\;}} \times {\rm{\;\vec a}}\)

\(\Rightarrow \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) = {\rm{\;}}0 + {\rm{\;\vec a\;}} \times {\rm{\;\vec b}} + {\rm{\;\vec a\;}} \times {\rm{\;\vec b}} - {\rm{\;}}0\)                \(\because \left( {{\rm{\vec a\;}} \times {\rm{\;\vec a}} = \;{\rm{\vec b\;}} \times {\rm{\;\vec b}} = 0} \right)\) 

\(\Rightarrow \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) = {\rm{\;}}2\;\left( {{\rm{\;\vec a\;}} \times {\rm{\;\vec b}}} \right)\)

∴ Option 3 is correct.

Concept:

If vectors \(\rm \vec a\;and\;\vec b\) are perpendicular then \(\rm \vec a \cdot \;\vec b = 0\)

Calculation:

Given: \(\rm 2\hat i - 5\hat j - \hat k\) and \(\rm -\hat i + 4 \hat j + λ\hat k\) are perpendicular

Let \(\rm \vec a = 2\hat i - 5\hat j - \hat k\) and \(\rm \vec b = -\hat i + 4 \hat j + λ\hat k\)

We know that, If vectors \(\rm \vec a\;and\;\vec b\) are perpendicular then \(\rm \vec a \cdot \;\vec b = 0\)

\(\rm \vec a \cdot \vec b = ( 2\hat i - 5\hat j - \hat k) \cdot (-\hat i + 4 \hat j + λ\hat k) = 0\)

⇒ -2 - 20 - λ = 0 

⇒ -22 - λ = 0 

∴ λ = -22

Concept:

Scaler triple product of the vectors:

The scaler triple product of the vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) is given by:

Coplaner vectors:

Three vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) are said to be coplaner if the scaler triple product \({\bf{\bar A}} \cdot \left[ {{\bf{\bar B}} \times {\bf{\bar C}}} \right] = 0.\)

Solution:

Let the given vectors be \({\rm{\bar A}} = {\rm{a}}\hat i + \;\hat j + {\rm{\;}}\hat k,\;\bar B = \;\hat i + \;b\hat j + \hat k\) and \({\rm{\bar C}} = {\rm{\;}}\hat i + \;\hat j + {\rm{c}}\hat k\).

It is given that the vectors are coplaner therefore the scaler triple product \({\rm{\bar A}} \cdot \left[ {{\rm{\bar B}} \times {\rm{\bar C}}} \right] = 0.\)

Therefore,

\(\left| {\begin{array}{*{20}{c}} {\rm{a}}&1&1\\ 1&{\rm{b}}&1\\ 1&1&{\rm{c}} \end{array}} \right| = 0\)

Perform the coloumn operation C₁ – C₂ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&1&1\\ {1 - {\rm{b}}}&{\rm{b}}&1\\ 0&1&{\rm{c}} \end{array}} \right| = 0\)

Now perform another coloum operation C₂ – C₃ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&0&1\\ {1 - {\rm{b}}}&{{\rm{b}} - 1}&1\\ 0&{1 - {\rm{c}}}&{\rm{c}} \end{array}} \right| = 0\)

Take (1 - a)(1 - b)(1 - c) common. Note that it is given that a,b,c ≠ 0 therefore this action is jstified.

\(\left( {1 - {\rm{a}}} \right)\left( {1 - {\rm{b}}} \right)\left( {1 - {\rm{c}}} \right)\left| {\begin{array}{*{20}{c}} { - 1}&0&{\frac{1}{{1 - {\rm{a}}}}}\\ 1&{ - 1}&{\frac{1}{{1 - {\rm{b}}}}}\\ 0&1&{\frac{{\rm{c}}}{{1 - {\rm{c}}}}} \end{array}} \right| = 0\)

Since a, b, c ≠ 0 therefore (1 - a)(1 - b)(1 - c) ≠ 0.Therefore the determinant has to be zero.

\( - 1\left( {\frac{{\rm-{c}}}{{1 - {\rm{c}}}} - {\rm{}}\frac{1}{{1 - {\rm{b}}}}} \right) + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{{\rm{c}}}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

Simplify the above expression as follows:

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} - {\rm{\;}}1 + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 1\)

Concept:

• If a, b, c are coplanar then [a b c] = 0
• Three vectors are permuted in the same cyclic order, the value of the scalar triple product remains the same. ⇒  [a b c] = [b c a] = [c a b]

Calculation:

Here, a, b, c are non-coplanar

To find: 2[a b c] + [b a c] =?

⇒ 2[a b c] + [b a c]

= 2 [a, b, c] - [a, b, c]                (∵ [b a c] = -[a b c])

= [a, b, c] 

Hence, option (2) is correct.

Concept:

If vectors a and b are orthogonal, then \(\rm \vec a.\vec b=0\)

Calculation: 

Here, \(\rm |\vec a+\vec b|=|\vec a-\vec b|\)

Squaring both sides we get, 

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

⇒4\(\rm \vec a.\vec b\)= 0

⇒\(\rm \vec a.\vec b\)= 0

So, Vectors a and b are orthogonal

As we know, if \(\rm \vec a.\vec b = 0\) then \(\rm \vec a \times \vec b \ne 0\)

So, only (1) is correct.

Hence, option (1) is correct.

Concept:

The cross product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \times \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ \;\hat n\) and \(|\vec a \times \vec b| = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ \;\)

The scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \)

If \(\vec a\) is a unit vector then \(|\vec a| = 1\)

Calculations:

Statement 1: The cross product of two unit vectors is always a unit vector.

Let \(\vec a\) and \(\vec b\) are two unit vectors.

i.e \(|\vec a| = 1 \ and \ |\vec b| = 1\)

As we know that, the cross product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \times \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ \;\hat n\) and \(|\vec a \times \vec b| = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ \;\)

⇒ \(|\vec a \times \vec b| = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ \; = sin θ \)

The range of sin θ is [-1, 1]

So, it is not necessarily true that the cross product of two unit vectors is always a unit vector.

Hence, statement 1 is false.

Statement 2: The dot product of two unit vectors is always unity.

Let \(\vec a\) and \(\vec b\) are two unit vectors.

i.e \(|\vec a| = 1 \ and \ |\vec b| = 1\)

As we know that, the scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \) 

⇒ \(|\vec a \cdot \;\vec b |= cos \ θ \)

The range of cos θ is [-1, 1].

So, it is not necessarily true that the dot product of two unit vectors is always a unit vector.

Hence, statement 2 is false.

Statement 3: The magnitude of sum of two unit vectors is always greater than the magnitude of their difference.

Let \(\vec a \ = \hat i \ and \ \ \vec b = \hat j\) 

As we can see that, the vectors  \(\vec a\) and \(\vec b\) are two unit vectors

⇒  \(|\hat i + \hat j| = \sqrt 2\) and \(|\hat i - \hat j| = \sqrt 2\)

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

So, statement 3 is also false.

Hence, the correct option is 4.

Concept:

Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

From above statement,

\(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \)

\( \Rightarrow \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = - \overrightarrow {{\rm{CA}}} \)

\(\therefore \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \vec 0\)

Only statement (1) is correct