M.C.Q (1 Marks) - Maths STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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MCQ 11 Mark

The position vectors of points $P$ and $Q$ are $\vec{p}$ and $\vec{q}$ respectively. The point $R$ divides line segment $P Q$ in the ratio $3: 1$ and $S$ is the mid-point of line segment $P R$. The position vector of $S$ is

A
$\frac{\vec{p}+3 \vec{q}}{4}$
B
$\frac{\vec{p}+3 \vec{q}}{8}$
C
$\frac{5 \vec{p}+3 \vec{q}}{4}$
D
$\frac{5 \vec{p}+3 \vec{q}}{8}$

Answer

Given, position vector of $P$ is $\overrightarrow{O P}=\vec{p}$ and position vector of $Q$ is $\overrightarrow{O P}=\vec{p}$, where $O$ is origin.
Point $R$ divides $P Q$ in ratio $3: 1$.
So, position vector of point $R$ is $\overrightarrow{O R}=\frac{3 \overrightarrow{O Q}+\overrightarrow{O P}}{4}=\frac{3 \vec{q}+\vec{p}}{4}$
Also, $S$ is the mid-point of $P R$.
So, $\overrightarrow{O S}=\frac{\overrightarrow{O P}+\overrightarrow{O R}}{2}=\frac{\vec{p}+\frac{3 \vec{q}+\vec{p}}{4}}{2}=\frac{5 \vec{p}+3 \vec{q}}{8}$

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MCQ 21 Mark

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=1,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=\sqrt{3}$, then the angle between $2 \vec{a}$ and $-\vec{b}$ is:

A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
C
$\frac{5 \pi}{6}$
D
$\frac{11 \pi}{6}$

Answer

We have, $|\vec{a}|=1,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=\sqrt{3}$
As, $\vec{a} \cdot \vec{b}=|a||b| \cos \theta$
$(2 \vec{a}) \cdot(-\vec{b})=|2 \vec{a}||-\vec{b}| \cos \theta$
$\Rightarrow-2(\vec{a} \cdot \vec{b})=2|\vec{a}||\vec{b}| \cos \theta \Rightarrow \cos \theta=\frac{-(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}=\frac{-\sqrt{3}}{2}$
$\therefore \quad$ Angle between $2 \vec{a}$ and $-\vec{b}=\pi-\frac{\pi}{6}$ or $\pi+\frac{\pi}{6}=\frac{5 \pi}{6}$ or $\frac{7 \pi}{6}$

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MCQ 31 Mark

The unit vector perpendicular to both vectors $\hat{i}+\hat{k}$ and $\hat{i}-\hat{k}$ is:

A
$2 \hat{j}$
B
$\hat{j}$
C
$\frac{\hat{i}-\hat{k}}{\sqrt{2}}$
D
$\frac{\hat{i}+\hat{k}}{\sqrt{2}}$

Answer

Let the required vector be $x \hat{i}+y \hat{j}+z \hat{k}$.
Then, $x^2+y^2+z^2=1$ .............(i)
Also, $(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{k})=0$
$\Rightarrow \quad x+z=0 ........(ii)$
And $(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-\hat{k})=0$
$\Rightarrow x - z =0 .........(iii)$
Solving (ii) and (iii), we get $x=z=0$
$\therefore \quad$ From (i), $y^2=1 \Rightarrow y= \pm 1$
So, required vector is $\pm \hat{j}$.

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MCQ 41 Mark

The vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{c}=-3 \hat{i}+4 \hat{j}+4 \hat{k}$ represents the sides of

A
an equilateral triangle
B
an obtuse-angled triangle
C
an isosceles triangle
D
a right-angled triangle

Answer

Given, $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}-3 \hat{j}-5 \hat{k}$
and $\vec{c}=-3 \hat{i}+4 \hat{j}+4 \hat{k}$, then
$
\begin{array}{l}
|\vec{a}|=\sqrt{4+1+1}=\sqrt{6},|\vec{b}|=\sqrt{1+9+25}=\sqrt{35} \text { and } \\
|\vec{c}|=\sqrt{9+16+16}=\sqrt{41}
\end{array}
$since,$|\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2$
$\therefore \quad$ It is a right - angled triangle.

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MCQ 51 Mark

Let $\vec{a}$ be any vector such that $|\vec{a}|=a$. The value of $|\vec{a} \times \hat{i}|^2+|\vec{a} \times \hat{j}|^2+|\vec{a} \times \hat{k}|^2$ is :

A
$a^2$
B
$2 a^2$
C
$3 a^2$
D
$0$

Answer

Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$, then $a^2=x^2+y^2+z^2$
Now, $\vec{a} \times \hat{i}=z \hat{j}-y \hat{k} \Rightarrow|\vec{a} \times \hat{i}|^2=y^2+z^2$
Similarly, $|\vec{a} \times \hat{j}|^2=x^2+z^2$ and $|\vec{a} \times \hat{k}|^2=x^2+y^2$
$\therefore \quad$ Required sum $=2 x^2+2 y^2+2 z^2=2 a^2$

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MCQ 61 Mark

For any two vectors $\vec{a}$ and $\vec{b}$, which of the following statements is always true?

A
$\vec{a} \cdot \vec{b} \geq|\vec{a}||\vec{b}|$
B
$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|$
✓
$\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|$
D
$\vec{a} \cdot \vec{b}<|\vec{a}||\vec{b}|$

Answer

Correct option: C.

$\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|$

$\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|$

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MCQ 71 Mark

If $\vec{a}=4 \hat{i}+6 \hat{j}$ and $\vec{b}=3 \hat{j}+4 \hat{k}$, then the vector form of the component of $\vec{a}$ along $\vec{b}$ is

A
$\frac{18}{5}(3 \hat{i}+4 \hat{k})$
B
$\frac{18}{25}(3 \hat{j}+4 \hat{k})$
C
$\frac{18}{5}(3 \hat{i}+4 \hat{k})$
D
$\frac{18}{25}(4 \hat{i}+6 \hat{j})$

Answer

$
\begin{array}{l}\text {Given, } \vec{a}=4 \hat{i}+6 \hat{j} \text { and } \vec{b}=3 \hat{j}+4 \hat{k} \\
\vec{a} \cdot \vec{b}=(4 \hat{i}+6 \hat{j}) \cdot(3 \hat{j}+4 \hat{k})=4 \times 0+6 \times 3+0 \times 4=18 \\
|\vec{b}|=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=5 \\
\therefore \quad|\vec{b}|^2=5^2=25
\end{array}$
Vector component of $\vec{a}$ along $\vec{b}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}=\frac{18}{25}(3 \hat{j}+4 \hat{k})$.

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MCQ 81 Mark

The value of $\lambda$ for which two vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $3 \hat{i}+\lambda \hat{j}+\hat{k}$ are perpendicular is

A
2
B
4
C
6
D
8

Answer

Dot product of two mutually perpendicular vectors is zero.
\[\begin{array}{l}
(2 \hat{i}-\hat{j}+2 \hat{k}) \cdot(3 \hat{i}+\lambda \hat{j}+\hat{k})=0 \\
\Rightarrow \quad 2 \times 3+(-1) \lambda+2 \times 1=0 \\
\Rightarrow \quad 6-\lambda+2=0 \Rightarrow \lambda=8
\end{array}\]

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MCQ 91 Mark

$A B C D$ is a rhombus whose diagonals intersects at $E$. Then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals to

A
$\overrightarrow{0}$
B
$\overrightarrow{ AD }$
C
$2 \overrightarrow{B D}$
D
$2 \overrightarrow{A D}$

Answer

Given, $A B C D$ is a rhombus whose diagonals bisect each other. $|\overrightarrow{E A}|=|\overrightarrow{E C}|$ and $|\overrightarrow{E B}|=|\overrightarrow{E D}|$ but since they are opposite to each other so they are of opposite signs.

$\Rightarrow \overrightarrow{E A}=-\overrightarrow{E C}$ and $\overrightarrow{E B}=-\overrightarrow{E D}$.
$\Rightarrow \quad \overrightarrow{E A}+\overrightarrow{E C}=\vec{O}$ ..........(i)
and $\overrightarrow{E B}+\overrightarrow{E D}=\vec{O}$ .......(ii)
Adding (i) and (ii), we get $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}=\vec{O}$.
$=2(1+\cos \theta)=4 \cos ^2 \frac{\theta}{2} \quad\left[\because 1+\cos \theta=2 \cos ^2 \frac{\theta}{2}\right]$
$\therefore|\vec{a}+\vec{b}|=2 \cos \frac{\theta}{2}$

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MCQ 101 Mark

If $(\hat{i}+\lambda \hat{j}) \times(5 \hat{i}+3 \hat{j}+\sigma \hat{k})=0$, what are the values of $\lambda$ and $\sigma$ ?

✓
$\lambda=\frac{3}{5}, \sigma=0$
B
$\lambda=\frac{5}{3}, \sigma=5$
C
$\lambda=3, \sigma=0$
D
(cannot be found as there are two unknowns and only one equation)

Answer

Correct option: A.

$\lambda=\frac{3}{5}, \sigma=0$

$\lambda=\frac{3}{5}, \sigma=0$

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MCQ 111 Mark

For which of these vectors is the projection on the $y$-axis zero?
(i) $2 \hat{j}$ (ii) $-5 \hat{k}$ (iii) $\hat{i}-4 \hat{k}$

A
only (i)
B
only (i) and (ii)
C
only (ii)
✓
only (ii) and (iii)

Answer

Correct option: D.

only (ii) and (iii)

(d) only (ii) and (i)

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MCQ 121 Mark

The projection of vector $\hat{i}$ on the vector $\hat{i}+\hat{j}+2 \hat{k}$ is:

A
$\frac{1}{\sqrt{6}}$
B
$\sqrt{6}$
C
$\frac{2}{\sqrt{6}}$
D
$\frac{3}{\sqrt{6}}$

Answer

Let, $\vec{a}=\hat{i}$ and $\vec{b}=\hat{i}+\hat{j}+2 \hat{k}$
We know that, projection of vector $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$
=\frac{\hat{i} \cdot(\hat{i}+\hat{j}+2 \hat{k})}{\sqrt{1^2+1^2+2^2}}=\frac{1}{\sqrt{6}}$

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MCQ 131 Mark

If $\vec{a}+\vec{b}=\hat{i}$ and $\vec{a}=2 \hat{i}-2 \hat{j}+2 \hat{k}$, then $|\vec{b}|$ equals:

A
$\sqrt{14}$
B
3
C
$\sqrt{12}$
D
$\sqrt{17}$

Answer

$\begin{array}{l}\text {Given, } \hat{a}+\hat{b}=\hat{i} \text { and } \vec{a}=2 \hat{i}-2 \hat{j}+2 \hat{k} \\ \Rightarrow \quad 2 \hat{i}-2 \hat{j}+2 \hat{k}+\vec{b}=\hat{i} \Rightarrow \vec{b}=\hat{i}-(2 \hat{i}-2 \hat{j}+2 \hat{k}) \\ \Rightarrow \quad-\hat{i}+2 \hat{j}-2 \hat{k} \\ \therefore \quad|\vec{b}|=\sqrt{(-1)^2+(2)^2+(-2)^2} \\ =\sqrt{1+4+4}=\sqrt{9}=3\end{array}$

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MCQ 141 Mark

If $A B C D$ is a parallelogram and $A C$ and $B D$ are its diagonals, then $\overrightarrow{A C}+\overrightarrow{B D}$ is:

A
$2 \overrightarrow{D A}$
B
$2 \overrightarrow{A B}$
C
$2 \overrightarrow{B C}$
D
$2 \overrightarrow{B D}$

Answer

Given, $A B C D$ is a parallelogram, then $A B \| C D$ and $B C \| D A$
$\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C} \quad$ [Triangle law of addition]
$\overrightarrow{B D}=\overrightarrow{B C}+\overrightarrow{C D}$ [Triangle law of addition]
$
\begin{array}{l}
\therefore \overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{B C}+\overrightarrow{C D} \\
=\overrightarrow{A B}+2 \overrightarrow{B C}-\overrightarrow{A B} \quad[\because \overrightarrow{A B}=-\overrightarrow{C D}] \\
=2 \overrightarrow{B C}
\end{array}$

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MCQ 151 Mark

The value of $(\hat{i} \times \hat{j}) \cdot \hat{j}+(\hat{j} \times \hat{i}) \cdot \hat{k}$ is:

A
2
B
$0$
C
1
D
-1

Answer

Since, $\hat{ i } \times \hat{ j }=\hat{ k }$ and $\hat{ j } \times \hat{ i }=-\hat{ k }$
$
\therefore \quad(\hat{i} \times \hat{j}) \cdot \hat{j}+(\hat{j} \times \hat{i}) \cdot \hat{k}=\hat{k} \cdot \hat{j}+(-\hat{k}) \cdot \hat{k}=0-\hat{k} \cdot \hat{k}=-1$

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MCQ 161 Mark

The magnitude of the vector $6 \hat{i}-2 \hat{j}+3 \hat{k}$ is

A
1
B
5
C
7
D
12

Answer

Given vector is $6 \hat{ i }-2 \hat{ j }+3 \hat{k}$
$
\begin{array}{l}
\therefore \text { Its magnitude }=\sqrt{6^2+(-2)^2+3^2} \\
=\sqrt{36+4+9}=\sqrt{49}=7 \text { units }
\end{array}$

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MCQ 171 Mark

The value of $p$ for which the vectors $2 \hat{i}+p \hat{j}+\hat{k}$ and $-4 \hat{i}-6 \hat{j}+26 \hat{k}$ are perpendicular to each other, is :

A
3
B
-3
C
$-\frac{17}{3}$
D
$\frac{17}{3} \quad$

Answer

Given vector are $2 \hat{i}+p \hat{j}+\hat{k}$ and $-4 \hat{i}-6 \hat{j}+26 \hat{k}$.
Since, the given vectors are perpendicular to each other.
$
\therefore \quad(2 \hat{i}+p \hat{j}+\hat{k}) \cdot(-4 \hat{i}-6 \hat{j}+26 \hat{k})=0
$
$\begin{array}{l}\Rightarrow \quad 2 \times(-4)+p \times(-6)+1 \times(26)=0 \\ \Rightarrow \quad-8-6 p+26=0 \Rightarrow-6 p+18=0 \\ \Rightarrow \quad-6 p=-18 \Rightarrow p=3 .\end{array}$

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MCQ 181 Mark

If $\vec{a}, \vec{b}$ and $(\vec{a}+\vec{b})$ are all unit vectors and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then the value of $\theta$ is :

A
$\frac{2 \pi}{3}$
B
$\frac{5 \pi}{6}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{6}$

Answer

Given, $|\vec{a}|=|\vec{b}|=|\vec{a}+\vec{b}|=1$
Now, $|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos \theta$
$
\begin{array}{l}
\Rightarrow 1^2=1^2+1^2+2 \cdot 1 \cdot 1 \cos \theta \\
\Rightarrow \cos \theta=\frac{-1}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-1}{2}\right) \Rightarrow \theta=\frac{2 \pi}{3}
\end{array}
$

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MCQ 191 Mark

A unit vector along the vector $4 \hat{i}-3 \hat{k}$ is

A
$\frac{1}{7}(4 \hat{i}-3 \hat{k})$
B
$\frac{1}{5}(4 \hat{i}-3 \hat{k})$
C
$\frac{1}{\sqrt{7}}(4 \hat{i}-3 \hat{k})$
D
$\frac{1}{\sqrt{5}}(4 \hat{i}-3 \hat{k})$

Answer

$\begin{array}{l}\text {Let } \vec{v}=4 \hat{i}-3 \hat{k} \\ \therefore|\vec{v}|=\sqrt{4^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5 \\ \text { Now, } \hat{v}=\text { unit vector along } \vec{v} \\ \quad=\frac{\vec{v}}{|\vec{v}|}=\frac{1}{5}(4 \hat{i}-3 \hat{k})\end{array}$

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MCQ 201 Mark

Two vectors $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ are collinear if

A
$a_1 b_1+a_2 b_2+a_3 b_3=0$
B
$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$
✓
$a_1=b_1, a_2=b_2, a_3=b_3$
D
$a_1+a_2+a_3=b_1+b_2+b_3$

Answer

Correct option: C.

$a_1=b_1, a_2=b_2, a_3=b_3$

$a_1=b_1, a_2=b_2, a_3=b_3$

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MCQ 211 Mark

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \geq 0$ only when

A
$0<\theta<\frac{\pi}{2}$
B
$0 \leq \theta \leq \frac{\pi}{2}$
C
$0<\theta<\pi$
D
$0 \leq \theta \leq \pi$

Answer

Given, $\vec{a} \cdot \vec{b} \geq 0 \Rightarrow|\vec{a}||\vec{b}| \cos \theta \geq 0$
Assuming $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$
$
\Rightarrow \cos \theta \geq 0 \quad[\because|\vec{a}| \geq 0,|\vec{b}| \geq 0] \Rightarrow \theta \in\left[0, \frac{\pi}{2}\right]
$

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MCQ 221 Mark

If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$, then $|\vec{a}-2 \vec{b}|$ is equal to

A
$\sqrt{2}$
B
$2 \sqrt{6}$
C
24
D
$2 \sqrt{2}$

Answer

$\begin{array}{l}\|\vec{a}-2 \vec{b}|^2=(\vec{a}-2 \vec{b}) \cdot(\vec{a}-2 \vec{b}) \\ =\vec{a} \cdot \vec{a}-4 \vec{a} \cdot \vec{b}+4 \vec{b} \cdot \vec{b}=|\vec{a}|^2-4 \vec{a} \cdot \vec{b}+4|\vec{b}|^2=4-16+36=24 \\ \therefore|\vec{a}-2 \vec{b}|=2 \sqrt{6}\end{array}$

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MCQ 231 Mark

The scalar projection of the vector $3 \hat{ i }-\hat{ j }-2 \hat{ k }$ on the vector $\hat{i}+2 \hat{j}-3 \hat{k}$ is

A
$\frac{7}{\sqrt{14}}$
B
$\frac{7}{14}$
C
$\frac{6}{13}$
D
$\frac{7}{2}$

Answer

Scalar projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ Scalar projection of $3 \hat{i}-\hat{j}-2 \hat{k}$ on vector $\hat{i}+2 \hat{j}-3 \hat{k}$
$
=\frac{(3 \hat{i}-\hat{j}-2 \hat{k}) \cdot(\hat{i}+2 \hat{j}-3 \hat{k})}{1 \hat{i}+2 \hat{j}-3 \hat{k} \mid}=\frac{7}{\sqrt{14}}$

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MCQ 241 Mark

If $\hat{i}, \hat{j}, \hat{k}$ are unit vectors along three mutually perpendicular directions, then

A
$\hat{i} \cdot \hat{j}=1$
B
$\hat{i} \times \hat{j}=1$
C
$\hat{ i } \cdot \hat{ k }=0$
D
$\hat{ i } \times \hat{ k }=0$

Answer

Since, $\hat{i}, \hat{j}, \hat{k}$ are mutually perpendicular to each other.
$
\therefore \hat{ i } \cdot \hat{ k }=0
$

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MCQ 251 Mark

$A B C D$ is a rhombus, whose diagonals intersect at $E$. Then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals

A
$\overrightarrow{0}$
B
$\overrightarrow{A D}$
C
$2 \overrightarrow{B C}$
D
$2 \overrightarrow{A D}$

Answer

$\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}=\overrightarrow{E A}+\overrightarrow{E B}-\overrightarrow{E A}-\overrightarrow{E B}$
[As diagonals of a rhombus bisect each other] $=\overrightarrow{0}$

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MCQ 261 Mark

if the projection of $\dot{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ on $\vec{b}=2 \hat{i}+\lambda \hat{k}$ is zero, then the value of $\lambda$ is

A
$0$
B
1
C
$\frac{-2}{3}$
D
$\frac{-3}{2}$

Answer

Here, $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+\lambda \hat{k}$
Since, projection of $\vec{a}$ on $\vec{b}=0$
$
\begin{array}{l}
\Rightarrow \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=0 \Rightarrow \frac{(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\lambda \hat{k})}{\sqrt{2^2+\lambda^2}}=0 \\
\Rightarrow \frac{2+3 \lambda}{\sqrt{4+\lambda^2}}=0 \Rightarrow 2+3 \lambda=0 \Rightarrow \lambda=-\frac{2}{3}
\end{array}$

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MCQ 271 Mark

The value of $p$ for which $p(\hat{i}+\hat{j}+\hat{k})$ is a unit vector is

A
$0$
B
$\frac{1}{\sqrt{3}}$
C
1
D
$\sqrt{3}$

Answer

Let $\vec{a}=(\hat{i}+\hat{j}+\hat{k})$
So, unit vector of $\vec{a}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{1+1+1}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
$\therefore \quad$ The value of $p$ is $\frac{1}{\sqrt{3}}$.

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MCQ 281 Mark

$(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2$ is equal to

A
1
B
$|\vec{a}|$
C
$-\vec{a}$
✓
$|\vec{a}|^2$

Answer

Correct option: D.

$|\vec{a}|^2$

(d) : Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k} \Rightarrow(\vec{a} \cdot \hat{i})^2=x^2$
Similarly, $(\vec{a} \cdot \hat{j})^2=y^2$ and $(\vec{a} \cdot \hat{k})^2=z^2$
$
\therefore \quad(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2=x^2+y^2+z^2=|\vec{a}|^2
$

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MCQ 291 Mark

If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}+2 \hat{k}$, then the value of $\vec{a} \cdot(\vec{b} \times \vec{c})$ is

A
-20
✓
-10
C
10
D
20

Answer

Correct option: B.

-10

(b): Here, $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}+2 \hat{k}$
Now, $\vec{b} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 3 & 1 & 2\end{array}\right|=3 \hat{i}+5 \hat{j}-7 \hat{k}$
$
\begin{array}{l}
\therefore \vec{a} \cdot(\vec{b} \times \vec{c})=(2 \hat{i}+\hat{j}+3 \hat{k}) \cdot(3 \hat{i}+5 \hat{j}-7 \hat{k}) \\
=2 \times 3+1 \times 5+3 \times(-7) \\
=6+5-21=-10
\end{array}
$

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MCQ 301 Mark

If $|\vec{a}|=3,|\vec{b}|=4$, then the value of $\lambda$ for which $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{a}-\lambda \vec{b}$, is

A
$\frac{9}{16}$
✓
$\frac{3}{4}$
C
$\frac{3}{2}$
D
$\frac{4}{3}$

Answer

Correct option: B.

$\frac{3}{4}$

(b) : Given that, $|\vec{a}|=3,|\vec{b}|=4$ and $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{a}-\lambda \vec{b}$.
$
\begin{array}{l}
\therefore(\vec{a}+\lambda \vec{b}) \cdot(\vec{a}-\lambda \vec{b})=0 \\
\Rightarrow \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b} \lambda+\lambda \vec{b} \cdot \vec{a}-\lambda^2 \vec{b} \cdot \vec{b}=0 \\
\Rightarrow|\vec{a}|^2-\lambda^2|\vec{b}|^2=0 \Rightarrow \lambda^2=\frac{|\vec{a}|^2}{|\vec{b}|^2} \Rightarrow \lambda=\frac{|\vec{a}|}{|\vec{b}|}=\frac{3}{4}
\end{array}
$

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MCQ 311 Mark

If $|\vec{a}|=4$ and $-3 \leq \lambda \leq 3$, then which of the following is the range of $|\lambda \vec{a}|$ ?
(i) $[0,8]$
(ii) $[-12,8]$
(iii) $[0,12]$

A
Only (i)
✓
Only (iii)
C
Only (i) and (ii)
D
All (i), (ii) and (iii)

Answer

Correct option: B.

Only (iii)

(b) : We have, $-3 \leq \lambda \leq 3 \Rightarrow|\lambda| \leq 3$
Now, $|\lambda||\vec{a}| \leq 3|\vec{a}| \Rightarrow|\lambda \vec{a}| \leq 12$
$\therefore \quad$ Range of $|\lambda \vec{a}|$ is $[0,12]$

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MCQ 321 Mark

$(\hat{i}+\hat{j}) \times(\hat{j}+\hat{k}) \cdot(\hat{k}+\hat{i})$ is equal to

A
0
B
1
C
2
D
-1

Answer

$
\begin{array}{l}
\text { (c) : }(\hat{i}+\hat{j}) \times(\hat{j}+\hat{k}) \cdot(\hat{k}+\hat{i})=(\hat{i} \times \hat{j}+\hat{i} \times \hat{k}+\hat{j} \times \hat{k}) \cdot(\hat{k}+\hat{i}) \\
=(\hat{k}-\hat{j}+\hat{i}) \cdot(\hat{k}+\hat{i})=\hat{k} \cdot \hat{k}+\hat{i} \cdot \hat{i} \quad(\because \hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0) \\
=|\hat{k}|^2+|\hat{i}|^2=1+1=2
\end{array}
$

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MCQ 331 Mark

If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$, then $|\vec{a} \times \vec{b}|$ is equal to

✓
$\sqrt{507}$
B
$\sqrt{506}$
C
$\sqrt{508}$
D
$\sqrt{509}$

Answer

Correct option: A.

$\sqrt{507}$

(a): We have, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2\end{array}\right|$
$
=\hat{i}(-2-15)-(-4-9) \hat{j}+(10-3) \hat{k}=-17 \hat{i}+13 \hat{j}+7 \hat{k}
$
Hence, $|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+(7)^2}=\sqrt{507}$

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MCQ 341 Mark

If $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=4 \hat{i}+4 \hat{j}-2 \hat{k}$ then find the angle between the vectors $\vec{a}$ and $\vec{b}$.

A
0
B
$\frac{\pi}{3}$
✓
$\frac{\pi}{2}$
D
$\frac{\pi}{4}$

Answer

Correct option: C.

$\frac{\pi}{2}$

(c) : We have, $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=4 \hat{i}+4 \hat{j}-2 \hat{k}$
Now, $\vec{a} \cdot \vec{b}=(2 \hat{i}-\hat{j}+2 \hat{k}) \cdot(4 \hat{i}+4 \hat{j}-2 \hat{k})$
$=8-4-4=0$. Therefore, $\vec{a} \cdot \vec{b}=0$
$
\Rightarrow \cos \theta=0
$
So, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.

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MCQ 351 Mark

If $A B C D$ is a rhombus, whose diagonals intersect at $E$, then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals

✓
$\overrightarrow{0}$
B
$\overrightarrow{A D}$
C
$2 \overrightarrow{B C}$
D
$2 \overrightarrow{A D}$

Answer

Correct option: A.

$\overrightarrow{0}$

(a) : $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$
$
=\overrightarrow{E A}+\overrightarrow{E B}-\overrightarrow{E A}-\overrightarrow{E B}
$ [As diagonals of a rhombus bisect each other]
$=\overrightarrow{0}$

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MCQ 361 Mark

Which of the following is the magnitude of the vector $\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ ?

A
6
✓
7
C
7.5
D
8.5

Answer

Correct option: B.

7

(b) : Here, $\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$
$\therefore \quad$ Its magnitude $=|\vec{a}|$
$
=\sqrt{3^2+(-2)^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7
$

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MCQ 371 Mark

The value of $p$ for which $p(\hat{i}+\hat{j}+\hat{k})$ is a unit vector is

A
0
✓
$\frac{1}{\sqrt{3}}$
C
1
D
$\sqrt{3}$

Answer

Correct option: B.

$\frac{1}{\sqrt{3}}$

(b) : Let $\vec{a}=(\hat{i}+\hat{j}+\hat{k})$
So, unit vector of $\vec{a}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{1+1+1}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
$\therefore \quad$ The value of $p$ is $\frac{1}{\sqrt{3}}$.

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MCQ 381 Mark

Let $\vec{a}$ and $\vec{b}$ are non-collinear. If $\vec{c}=(x-2) \vec{a}+\vec{b}$ and $\vec{d}=(2 x+1) \vec{a}-\vec{b}$ are collinear, then find the value of $x$.

A
$\frac{2}{3}$
B
$\frac{-1}{3}$
C
$\frac{-2}{3}$
✓
$\frac{1}{3}$

Answer

Correct option: D.

$\frac{1}{3}$

(d): We have, $\vec{c}=(x-2) \vec{a}+\vec{b}, \vec{d}=(2 x+1) \vec{a}-\vec{b}$ are collinear, then $\vec{c}=m \vec{d}$ where $m$ is any scalar.
$
\begin{array}{l}
\Rightarrow \quad(x-2) \vec{a}+\vec{b}=m((2 x+1) \vec{a}-\vec{b}) \\
\Rightarrow \quad-m=1 \Rightarrow m=-1 \\
\text { and } m(2 x+1)=x-2 \Rightarrow-2 x-1=x-2 \Rightarrow x=\frac{1}{3}
\end{array}
$

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MCQ 391 Mark

If $\vec{u}=\hat{i}+2 \hat{j}, \vec{v}=-2 \hat{i}+\hat{j}$ and $\vec{w}=4 \hat{i}+3 \hat{j}$, then find scalars $x$ and $y$ such that $\vec{w}=x \vec{u}+y \vec{v}$.

A
$x=4, y=-2$
✓
$x=2, y=-1$
C
$x=3, y=5$
D
$x=-5, y=2$

Answer

Correct option: B.

$x=2, y=-1$

(b): We have, $\vec{w}=x \vec{u}+y \vec{v}$
$
\begin{array}{l}
\Rightarrow \quad 4 \hat{i}+3 \hat{j}=x(\hat{i}+2 \hat{j})+y(-2 \hat{i}+\hat{j}) \\
\Rightarrow \quad(x-2 y-4) \hat{i}+(2 x+y-3) \hat{j}=\overrightarrow{0} \\
\Rightarrow \quad x-2 y-4=0 \text { and } 2 x+y-3=0 \\
\Rightarrow \quad x=2 \text { and } y=-1
\end{array}
$

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MCQ 401 Mark

If $|\vec{a}-\vec{b}|=|\vec{a}|=|\vec{b}|=1$, then the angle between $\vec{a}$ and $\vec{b}$ is

✓
$\frac{\pi}{3}$
B
$\frac{3 \pi}{4}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{6}$

Answer

Correct option: A.

$\frac{\pi}{3}$

(a) : Given, $|\vec{a}-\vec{b}|=|\vec{a}|=|\vec{b}|=1$
$
\Rightarrow|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} \Rightarrow 1=1+1-2|\vec{a}||\vec{b}| \cos \theta
$
(Here $\theta$ is angle between $\vec{a}$ and $\vec{b}$ )
$
\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
$

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MCQ 411 Mark

Which of the following is the vector in the direction of the vector $\hat{i}-2 \hat{j}+2 \hat{k}$ that has magnitude 9?

A
$\hat{i}-2 \hat{j}+2 \hat{k}$
B
$\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}$
C
$3(\hat{i}-2 \hat{j}+2 \hat{k})$
D
$9(\hat{i}-2 \hat{j}+2 \hat{k})$

Answer

$
\begin{array}{l}
\text { (c) : Let } \vec{a}=\hat{i}-2 \hat{j}+2 \hat{k} \\
\therefore|\vec{a}|=\sqrt{1+4+4}=\sqrt{9}=3 \\
\therefore \quad \text { Required vector }=\frac{9(\hat{i}-2 \hat{j}+2 \hat{k})}{3}=3(\hat{i}-2 \hat{j}+2 \hat{k})
\end{array}
$

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MCQ 421 Mark

If the angle between $\hat{i}+\hat{k}$ and $\hat{i}+\hat{j}+a \hat{k}$ is $\frac{\pi}{3}$ then the value of $a$ is

A
2
B
-4 or 0
✓
0
D
2 or -2

Answer

Correct option: C.

0

(c) : We have, $\cos \frac{\pi}{3}=\frac{(\hat{i}+\hat{k}) \cdot(\hat{i}+\hat{j}+a \hat{k})}{\sqrt{2} \sqrt{1+1+a^2}}$
$
\begin{array}{l}
\Rightarrow \frac{1}{2}=\frac{1+a}{\sqrt{2} \sqrt{2+a^2}} \\
\Rightarrow \frac{1}{4}=\frac{(1+a)^2}{2\left(2+a^2\right)} & ...(i) \\
\Rightarrow 2+a^2=2\left(1+a^2+2 a\right) \Rightarrow a^2+4 a=0 \Rightarrow a=0,-4
\end{array}
$
But $a=-4$ does not satisfy equation (i), Hence, the required value of $a$ is 0 .

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MCQ 431 Mark

If $A$ and $B$ are the points $(-3,4,-8)$ and $(5,-6,4)$ respectively, then find the ratio in which $y z$-plane divides $\overrightarrow{A B}$.

A
$5: 2$
B
$7: 5$
✓
$3: 5$
D
$5: 3$

Answer

Correct option: C.

$3: 5$

(c) : Let $\vec{a}=-3 \hat{i}+4 \hat{j}-8 \hat{k}, \vec{b}=5 \hat{i}-6 \hat{j}+4 \hat{k}$
Let $C(\vec{c})$ be the point in $y z$-plane which divides $\overrightarrow{A B}$ in the ratio $r: 1$ internally.
Then, $0=\frac{5 r-3}{r+1} \quad(\because$ In $y z$-plane, $x=0)$
$\Rightarrow \quad 5 r-3=0 \Rightarrow r=\frac{3}{5}$
Thus required ratio is $3: 5$

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MCQ 441 Mark

Let $\vec{a}$ and $\vec{b}$ are unit vectors enclosing an angle $\theta$ and $|\vec{a}+\vec{b}|<1$. Which of the following is true?
(i) $\theta=\frac{\pi}{2}$
(ii) $\theta<\frac{\pi}{3}$
(iii) $\pi \geq \theta>\frac{2 \pi}{3}$
(iv) $\cos \theta<-\frac{1}{2}$

A
Only (i)
B
Only (ii)
C
Only (iii) and (iv)
D
None of these

Answer

$
\begin{array}{l}
\text { (c) : }|\vec{a}+\vec{b}|<1 \Rightarrow|\vec{a}+\vec{b}|^2<1 \\
\Rightarrow|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}<1 \Rightarrow 1+1+2 \vec{a} \cdot \vec{b}<1 \quad[\because|\vec{a}|=|\vec{b}|=1] \\
\Rightarrow \vec{a} \cdot \vec{b}<-\frac{1}{2} \Rightarrow|\vec{a}||\vec{b}| \cos \theta<-\frac{1}{2} \\
\Rightarrow 1 \times 1 \times \cos \theta<-\frac{1}{2} \Rightarrow \cos \theta<-\frac{1}{2} \\
\Rightarrow-1 \leq \cos \theta<-\frac{1}{2} \Rightarrow \pi \geq \theta>\frac{2 \pi}{3}
\end{array}
$

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MCQ 451 Mark

Find the value of $\lambda$ so that the vectors $2 \hat{i}-4 \hat{j}+\hat{k}$ and $4 \hat{i}-8 \hat{j}+\lambda \hat{k}$ are perpendicular.

A
20
✓
-40
C
40
D
-20

Answer

Correct option: B.

-40

(b): The given vectors are perpendicular if their dot product vanishes, i.e.,
$
\begin{aligned}
& (2 \hat{i}-4 \hat{j}+\hat{k}) \cdot(4 \hat{i}-8 \hat{j}+\lambda \hat{k})=0 \\
\Rightarrow & 8+32+\lambda=0 \Rightarrow \lambda=-40
\end{aligned}
$

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MCQ 461 Mark

The vectors from origin to the points $A$ and $B$ are $\vec{a}=2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$, respectively, then the area of triangle $O A B$ (in sq. units) is

A
$\sqrt{340}$
B
$\sqrt{325}$
C
$\sqrt{229}$
✓
$\frac{1}{2} \sqrt{229}$

Answer

Correct option: D.

$\frac{1}{2} \sqrt{229}$

(d) : $\vec{a}=2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$
$
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 2 \\
2 & 3 & 1
\end{array}\right|=-9 \hat{i}+2 \hat{j}+12 \hat{k}
$
Area of $\triangle O A B=\frac{1}{2}|\vec{a} \times \vec{b}|$
$
=\frac{1}{2} \sqrt{81+4+144}=\frac{1}{2} \sqrt{229} \text { sq. units }
$

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MCQ 471 Mark

If $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+p \hat{j}+q \hat{k})=\overrightarrow{0}$, then which of the following is true?

A
$p=6, q=27$
✓
$p=3, q=\frac{27}{2}$
C
$p=6, q=\frac{27}{2}$
D
$p=3, q=27$

Answer

Correct option: B.

$p=3, q=\frac{27}{2}$

(b) : Given,
$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & p & q\end{array}\right|=0$
$
\begin{array}{ll}
\Rightarrow & \hat{i}(6 q-27 p)-\hat{j}(2 q-27)+\hat{k}(2 p-6)=0 \\
\Rightarrow & 6 q-27 p=0,2 q-27=0 \text { and } 2 p-6=0 \\
\Rightarrow & q=\frac{27}{2} \text { and } p=3 .
\end{array}
$

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MCQ 481 Mark

The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4 , respectively and $\vec{a} \cdot \vec{b}=2 \sqrt{3}$ is

A
$\frac{\pi}{6}$
✓
$\frac{\pi}{3}$
C
$\frac{\pi}{2}$
D
$\frac{5 \pi}{2}$

Answer

Correct option: B.

$\frac{\pi}{3}$

(b) : We have $\vec{a} \cdot \vec{b}=2 \sqrt{3},|\vec{a}|=\sqrt{3},|\vec{b}|=4$
Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.
$
\begin{array}{ll}
\therefore & \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\
\Rightarrow & 2 \sqrt{3}=\sqrt{3} \cdot 4 \cdot \cos \theta \quad \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
$

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MCQ 491 Mark

Area of a parallelogram whose adjacent sides are represented by the vectors $2 \hat{i}-3 \hat{k}$ and $4 \hat{j}+2 \hat{k}$ is

✓
$4 \sqrt{14}$ sq. units
B
$2 \sqrt{7}$ sq. units
C
$4 \sqrt{7}$ sq. units
D
$4 \sqrt{19}$ sq. units

Answer

Correct option: A.

$4 \sqrt{14}$ sq. units

(a) : Let $\vec{a}=2 \hat{i}-3 \hat{k}$ and $\vec{b}=4 \hat{j}+2 \hat{k}$
The area of a parallelogram with $\vec{a}$ and $\vec{b}$ as its adjacent sides is given by $|\vec{a} \times \vec{b}|$.
Now, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -3 \\ 0 & 4 & 2\end{array}\right|=12 \hat{i}-4 \hat{j}+8 \hat{k}$
$
\begin{array}{l}
\therefore \quad|\vec{a} \times \vec{b}|=\sqrt{(12)^2+(-4)^2+(8)^2}=\sqrt{144+16+64} \\
\quad=\sqrt{224}=4 \sqrt{14} \text { sq. units. }
\end{array}
$

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MCQ 501 Mark

If $|\vec{a}|=10,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$, then the value of $|\vec{a} \times \vec{b}|$ is

A
5
B
10
C
14
✓
16

Answer

Correct option: D.

16

(d) : $|\vec{a}|=10,|\vec{b}|=2, \vec{a} \cdot \vec{b}=12$
We know that, $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
$
\begin{array}{l}
\Rightarrow 12=10 \times 2 \cos \theta \Rightarrow \cos \theta=\frac{3}{5} \\
\therefore \sin \theta=\frac{4}{5}
\end{array}
$
Now, $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=10 \times 2 \times \frac{4}{5}=16$

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M.C.Q (1 Marks) - Maths STD 12 Science Questions - Vidyadip