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

Concept:

If $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors parallel to each other then $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=0$

Calculation:

Given:

 $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}$ and $3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}}$ are parallel vectors,

Therefore, $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}=\lambda (3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}})$

Equating the coefficient of $\overrightarrow{\mathrm{i}},\overrightarrow{\mathrm{j}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{k}}$

⇒ 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:

$\overrightarrow{AB}=2\overrightarrow{FC}$

⇒ $\overrightarrow{AB}=-2\overrightarrow{CF}$

⇒ $n=-\frac{1}{2}$

Also,

$\overrightarrow{AD}=2\overrightarrow{BC}\Rightarrow m=2$

Now,

$mn=2(-\frac{1}{2})=-1$

∴ The final value of  is -1.

Calculation

Given

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

Any point on the vector $\overrightarrow{r}$ can be taken as,

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

$\overrightarrow{PQ}=\{-3\mu -2,\mu -3,5\mu -4\}$

Now, the $\overrightarrow{PQ}$ must be perpendicular to the normal for the given plane.

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

$\Rightarrow -3\mu -2-4\mu +12+15\mu -12=0$

$\Rightarrow 8\mu =2$

$\Rightarrow \mu ={\displaystyle \frac{1}{4}}$

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.

⇒ $\frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda }$

∴  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 $\overrightarrow{a}$ is given by, $\hat{a}$ = $\frac{\overrightarrow{a}}{|\overrightarrow{a}|}$ where $|\overrightarrow{a}|$ is the magnitude of the vector.

Calculation:

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

∴ $\overrightarrow{R}$ = $\overrightarrow{r_1}$ + $\overrightarrow{r_2}$ = $-\overrightarrow{i}+2\overrightarrow{j}$

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

⇒ Unit vector, $\widehat{R_1}$ = $\frac{\overrightarrow{R}}{|\overrightarrow{R}|}$

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

∴ The unit vector which is paralleled to the addition of two vectors is $1/\sqrt{5}(-\overrightarrow{i}+2\overrightarrow{j})$.

The correct answer is Option 3.

Concept:

If $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors parallel to each other then $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=0$

Calculation:

Given:

 $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}$ and $3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}}$ are parallel vectors,

Therefore, $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}=\lambda (3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}})$

Equating the coefficient of $\overrightarrow{\mathrm{i}},\overrightarrow{\mathrm{j}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{k}}$

⇒ 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 $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors parallel to each other then $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=0$

Calculation:

Given:

 $\mathrm{x}\overrightarrow{\mathrm{i}}-2\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}$ and $2\overrightarrow{\mathrm{i}}-4\overrightarrow{\mathrm{j}}+\mathrm{y}\overrightarrow{\mathrm{k}}$ are parallel vectors,

Therefore, $\mathrm{x}\overrightarrow{\mathrm{i}}-2\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}=\lambda (2\overrightarrow{\mathrm{i}}-4\overrightarrow{\mathrm{j}}+\mathrm{y}\overrightarrow{\mathrm{k}})$

Equating the coefficient of $\overrightarrow{\mathrm{i}},\overrightarrow{\mathrm{i}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{k}}$

⇒ 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.

⇒ $\frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda }$

∴  λ = $\frac{2}{3}$

The value of λ is $\frac{2}{3}$.

The correct answer is option 1.

Concept:

Unit vector parallel to $\overrightarrow{a}=\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

Calculation:

Let $\overrightarrow{a}$ = -2î + 3ĵ

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

∴ Unit vector parallel to $\overrightarrow{a}=\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

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

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

Calculation:

$\overrightarrow{AB}=2\overrightarrow{FC}$

⇒ $\overrightarrow{AB}=-2\overrightarrow{CF}$

⇒ $n=-\frac{1}{2}$

Also,

$\overrightarrow{AD}=2\overrightarrow{BC}\Rightarrow m=2$

Now,

$mn=2(-\frac{1}{2})=-1$

∴ The final value of  is -1.

Concept:

If $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors parallel to each other then $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=0$

Calculation:

Given:

 $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}$ and $3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}}$ are parallel vectors,

Therefore, $\overrightarrow{\mathrm{i}}-\mathrm{a}\overrightarrow{\mathrm{j}}+5\overrightarrow{\mathrm{k}}=\lambda (3\overrightarrow{\mathrm{i}}-6\overrightarrow{\mathrm{j}}+\mathrm{b}\overrightarrow{\mathrm{k}})$

Equating the coefficient of $\overrightarrow{\mathrm{i}},\overrightarrow{\mathrm{j}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{k}}$

⇒ 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 $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are two vectors parallel to each other then $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=0$

Calculation:

Given:

 $\mathrm{x}\overrightarrow{\mathrm{i}}-2\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}$ and $2\overrightarrow{\mathrm{i}}-4\overrightarrow{\mathrm{j}}+\mathrm{y}\overrightarrow{\mathrm{k}}$ are parallel vectors,

Therefore, $\mathrm{x}\overrightarrow{\mathrm{i}}-2\overrightarrow{\mathrm{j}}+3\overrightarrow{\mathrm{k}}=\lambda (2\overrightarrow{\mathrm{i}}-4\overrightarrow{\mathrm{j}}+\mathrm{y}\overrightarrow{\mathrm{k}})$

Equating the coefficient of $\overrightarrow{\mathrm{i}},\overrightarrow{\mathrm{i}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{k}}$

⇒ 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.

⇒ $\frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda }$

∴  λ = $\frac{2}{3}$

The value of λ is $\frac{2}{3}$.

The correct answer is option 1.

Concept:

Unit vector parallel to $\overrightarrow{a}=\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

Calculation:

Let $\overrightarrow{a}$ = -2î + 3ĵ

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

∴ Unit vector parallel to $\overrightarrow{a}=\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

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

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

Given:

$\overrightarrow{A}$ =   î + ĵ + k̂,

$\overrightarrow{B}$  = 2î + 3ĵ,

$\overrightarrow{C}$  = 3î + 5ĵ - 2k̂ and,

$\overrightarrow{D}$ =  k̂ - ĵ

Concept:

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

 $\overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}}$, where λ = some constant

Calculation:

Consider a vector $\overrightarrow{O}$  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.