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

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:

\( \vec{AB} = 2\vec{FC} \)

⇒ \( \vec{AB} = -2\vec{CF} \)

⇒ \( n = -\frac{1}{2} \)

Also,

\( \vec{AD} = 2\vec{BC} \Rightarrow m = 2 \)

Now,

\( mn = 2 \left( -\frac{1}{2} \right) = -1 \)

∴ The final value of  is -1.

Calculation

Given

\(\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(-3\hat{i}+\hat{j}+5\hat{k})\)

Any point on the vector \(\vec{r}\) can be taken as,

\( Q \equiv \{ (1-3\mu ),(\mu -1),(5\mu +2)\} \) gives

\(\vec { { P }{ Q } } =\{ -3\mu -2,\mu -3,5\mu -4\} \)

Now, the \(\vec{PQ}\) must be perpendicular to the normal for the given plane.

⇒ \(1(-3\mu -2)-4(\mu -3)+3(5\mu -4)=0\ \)

\(⇒ -3\mu -2-4\mu +12+15\mu -12=0\ \)

\(⇒ 8\mu =2\)

\(⇒ \mu =\dfrac {1}{4}\)

Hence option (1) is correct

Concept:

If two vectors  a₁î + b₁ĵ + c₁k̂ and a₂î + b₂ĵ + c₂k̂ are parallel then

​\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Calculation:

Given two vectors are 3î − 6ĵ + k̂ and 2î − 4ĵ + λk̂.

Given two vectors are parallel.

⇒ \(\frac{3}{2}=\frac{-6}{-4}=\frac{1}{λ }\)

∴  λ = \(\frac{2}{3}\)

The value of λ is \(\frac{2}{3}\).

The correct answer is option 1.

Concept:

The unit vector in the direction of \(\vec{a}\) is given by, \(\hat{a}\) = \(\frac{\vec{a}}{|\vec{a}|}\) where \(|\vec{a}|\) is the magnitude of the vector.

Calculation:

Given, \(\overrightarrow{r_1}\) = \(3\vec{i}-2\vec{j}\) and \(\overrightarrow{r_2}\) = – \(4\vec{i}+4\vec{j}\)

∴ \(\vec{R}\) = \(\vec{r_1}\) + \(\vec{r_2}\) = \(-\vec{i}+2\vec{j}\)

⇒ \(|\vec{R}|\) = \(\sqrt{(-1)^2+2^2}\) = \(\sqrt{5}\)

⇒ Unit vector, \(\hat{R_1}\) = \(\frac{\vec{R}}{|\vec{R}|}\)

= \(\frac{1}{\sqrt{5}}\)(\(-\vec{i}+2\vec{j}\))

∴ The unit vector which is paralleled to the addition of two vectors is \(1/\sqrt5(-\vec{i}+2\vec{j})\).

The correct answer is Option 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

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:

 \(\rm x\vec{i} - 2\vec{j} + 3\vec{k}\) and \(\rm 2\vec{i} - 4\vec{j} + y\vec{k}\) are parallel vectors,

Therefore, \(\rm x\vec{i} - 2\vec{j} + 3\vec{k} = λ (\rm 2\vec{i} - 4\vec{j} + y\vec{k})\)

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

⇒ x = 2λ                    .... (1)

⇒ -2 = -4λ 

∴ λ = 1/2

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

x = 2 × (1/2)

So, x = 1

Concept:

If two vectors  a₁î + b₁ĵ + c₁k̂ and a₂î + b₂ĵ + c₂k̂ are parallel then

​\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Calculation:

Given two vectors are 3î − 6ĵ + k̂ and 2î − 4ĵ + λk̂.

Given two vectors are parallel.

⇒ \(\frac{3}{2}=\frac{-6}{-4}=\frac{1}{λ }\)

∴  λ = \(\frac{2}{3}\)

The value of λ is \(\frac{2}{3}\).

The correct answer is option 1.

Concept:

Unit vector parallel to \(\vec{a}=\hat{a}=\frac{\vec{a}}{\left |\vec{a} \right |}\)

Calculation:

Let \(\vec{a}\) = -2î + 3ĵ

\(\Rightarrow \left | \vec{a} \right |=\sqrt{(-2)^2+3^2}=\sqrt{13}\)

∴ Unit vector parallel to \(\vec{a}=\hat{a}=\frac{\vec{a}}{\left |\vec{a} \right |}\)

\(\Rightarrow \frac{1}{\sqrt{13}}(-2\hat{i}+3\hat{j})\)

\(\Rightarrow \frac{-2\hat{i}}{\sqrt{13}}+\frac{3\hat{j}}{\sqrt{13}}\)

Calculation:

\( \vec{AB} = 2\vec{FC} \)

⇒ \( \vec{AB} = -2\vec{CF} \)

⇒ \( n = -\frac{1}{2} \)

Also,

\( \vec{AD} = 2\vec{BC} \Rightarrow m = 2 \)

Now,

\( mn = 2 \left( -\frac{1}{2} \right) = -1 \)

∴ The final value of  is -1.

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

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:

 \(\rm x\vec{i} - 2\vec{j} + 3\vec{k}\) and \(\rm 2\vec{i} - 4\vec{j} + y\vec{k}\) are parallel vectors,

Therefore, \(\rm x\vec{i} - 2\vec{j} + 3\vec{k} = λ (\rm 2\vec{i} - 4\vec{j} + y\vec{k})\)

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

⇒ x = 2λ                    .... (1)

⇒ -2 = -4λ 

∴ λ = 1/2

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

x = 2 × (1/2)

So, x = 1

Concept:

If two vectors  a₁î + b₁ĵ + c₁k̂ and a₂î + b₂ĵ + c₂k̂ are parallel then

​\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Calculation:

Given two vectors are 3î − 6ĵ + k̂ and 2î − 4ĵ + λk̂.

Given two vectors are parallel.

⇒ \(\frac{3}{2}=\frac{-6}{-4}=\frac{1}{λ }\)

∴  λ = \(\frac{2}{3}\)

The value of λ is \(\frac{2}{3}\).

The correct answer is option 1.

Concept:

Unit vector parallel to \(\vec{a}=\hat{a}=\frac{\vec{a}}{\left |\vec{a} \right |}\)

Calculation:

Let \(\vec{a}\) = -2î + 3ĵ

\(\Rightarrow \left | \vec{a} \right |=\sqrt{(-2)^2+3^2}=\sqrt{13}\)

∴ Unit vector parallel to \(\vec{a}=\hat{a}=\frac{\vec{a}}{\left |\vec{a} \right |}\)

\(\Rightarrow \frac{1}{\sqrt{13}}(-2\hat{i}+3\hat{j})\)

\(\Rightarrow \frac{-2\hat{i}}{\sqrt{13}}+\frac{3\hat{j}}{\sqrt{13}}\)

Given:

\(\vec{A}\) =   î + ĵ + k̂,

\(\vec{B}\)  = 2î + 3ĵ,

\(\vec{C}\)  = 3î + 5ĵ - 2k̂ and,

\(\vec{D}\) =  k̂ - ĵ

Concept:

If two vectors \(\vec{a}\) and \(\vec{b}\) are collinear then vectors can be written as a linear expression of other vectors:

 \(\rm \vec a=λ\vec b\), where λ = some constant

Calculation:

Consider a vector \(\vec{O}\)  with position vector 0î + 0ĵ + 0k̂.

AB = OB - OA

AB = 2î + 3ĵ - î - ĵ - k̂

AB = î + 2ĵ - k̂       -----(1)

Similarly, 

CD = OD - OC

CD = k̂ - ĵ - 3î - 5ĵ + 2k̂

CD = - 3î - 6ĵ + 3k̂ 

CD = -3(î + 2ĵ - k̂)       -----(2)

From equatiion (1) & (2), we can see

CD = -3(AB)

So, they are collinear vector

∴ AB and CD are parallel.