M.C.Q (1 Marks)

MCQ 511 Mark

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined to $x$-axis at angles $30^{\circ}$ and $120^{\circ}$ respectively, then $|\vec{a}+\vec{b}|$ equals

A
$\sqrt{\frac{2}{3}}$
✓
$\sqrt{2}$
C
$\sqrt{3}$
D
2

Answer

Correct option: B.

$\sqrt{2}$

(b) : Clearly, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.
$
\Rightarrow \quad \vec{a} \cdot \vec{b}=0
$
$
\begin{array}{l}
\therefore|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1+1+0=2 \\
\Rightarrow|\vec{a}+\vec{b}|=\sqrt{2}
\end{array}
$

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MCQ 521 Mark

The direction ratios of the vector $3 \vec{a}+2 \vec{b}$, where $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ are

A
$7,5,4$
✓
$7,-5,4$
C
$-7,5,4$
D
$7,5,-4$

Answer

Correct option: B.

$7,-5,4$

(b) : We have, $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}, \vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$
$
\begin{array}{l}
\therefore 3 \vec{a}+2 \vec{b}=3(\hat{i}+\hat{j}-2 \hat{k})+2(2 \hat{i}-4 \hat{j}+5 \hat{k}) \\
=(3 \hat{i}+3 \hat{j}-6 \hat{k})+(4 \hat{i}-8 \hat{j}+10 \hat{k})=7 \hat{i}-5 \hat{j}+4 \hat{k}
\end{array}
$
$\therefore \quad$ The direction ratios of the vector $3 \vec{a}+2 \vec{b}$ are $7,-5,4$.

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MCQ 531 Mark

Which of these are the direction cosines of the vector $-2 \hat{i}+\hat{j}-5 \hat{k}$ ?

A
$\left(\frac{2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{5}{\sqrt{30}}\right)$
✓
$\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)$
C
$\left(-\frac{2}{\sqrt{30}},-\frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)$
D
None of these

Answer

Correct option: B.

$\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)$

(b): We have, $\vec{a}=-2 \hat{i}+\hat{j}-5 \hat{k}$
Direction cosines of the given vector are
$
\left.\begin{array}{rl}
\left(\frac{-2}{\sqrt{(-2)^2+(1)^2+(-5)^2}}, \frac{1}{\sqrt{(-2)^2+(1)^2+(-5)^2}}\right. ,\frac{-5}{\sqrt{(-2)^2+(1)^2+(-5)^2}}
\end{array}\right)
$
$
=\left(\frac{-2}{\sqrt{4+1+25}}, \frac{1}{\sqrt{4+1+25}}, \frac{-5}{\sqrt{4+1+25}}\right)
$
$\therefore$ Direction cosines are $\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)$

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MCQ 541 Mark

Find the value of $\lambda$ for which the vectors $3 \hat{i}-6 \hat{j}+\hat{k}$ and $2 \hat{i}-4 \hat{j}+\lambda \hat{k}$ are parallel.

✓
$\frac{2}{3}$
B
$\frac{-3}{2}$
C
$\frac{-2}{3}$
D
$\frac{3}{2}$

Answer

Correct option: A.

$\frac{2}{3}$

(a) : $\vec{a}=3 \hat{i}-6 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}-4 \hat{j}+\lambda \hat{k}$
Since, $\vec{a}$ and $\vec{b}$ are parallel $\quad \therefore \quad \vec{a} \times \vec{b}=\overrightarrow{0}$
$
\begin{aligned}
\Rightarrow & \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -6 & 1 \\
2 & -4 & \lambda
\end{array}\right|=\overrightarrow{0} \\
\Rightarrow & (-6 \lambda+4) \hat{i}-(3 \lambda-2) \hat{j}+(-12+12) \hat{k}=\overrightarrow{0} \\
\Rightarrow & (-6 \lambda+4) \hat{i}+(2-3 \lambda) \hat{j}=0 \hat{i}+0 \hat{j}
\end{aligned}
$
Comparing coefficients of $\hat{i}$ and $\hat{j}$, we get $-6 \lambda+4=0$ and $2-3 \lambda=0 \Rightarrow \lambda=2 / 3$

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MCQ 551 Mark

Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$, $\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.

✓
$-4 \hat{j}-\hat{k}$
B
$-\hat{i}-4 \hat{j}-\hat{k}$
C
$4 \hat{j}+\hat{k}$
D
$\hat{i}-4 \hat{j}$

Answer

Correct option: A.

$-4 \hat{j}-\hat{k}$

(a): The given vectors are
$
\begin{array}{l}
\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}, \vec{c}=\hat{i}-6 \hat{j}-7 \hat{k} \\
\therefore \quad \text { Required sum }=\vec{a}+\vec{b}+\vec{c} \\
=(\hat{i}-2 \hat{j}+\hat{k})+(-2 \hat{i}+4 \hat{j}+5 \hat{k})+(\hat{i}-6 \hat{j}-7 \hat{k}) \\
=-4 \hat{j}-\hat{k}
\end{array}
$

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MCQ 561 Mark

The magnitude of each of the two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{9}{2}$, is

A
2
✓
3
C
4
D
5

Answer

Correct option: B.

(b) : Given, $|\vec{a}|=|\vec{b}|, \theta=60^{\circ}$ and $\vec{a} \cdot \vec{b}=\frac{9}{2}$
Now, $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$
\begin{array}{l}
\Rightarrow \cos 60^{\circ}=\frac{9 / 2}{|\vec{a}|^2} \Rightarrow \frac{1}{2}=\frac{9 / 2}{|\vec{a}|^2} \\
\Rightarrow|\vec{a}|^2=9 \Rightarrow|\vec{a}|=3 \therefore|\vec{a}|=|\vec{b}|=3
\end{array}
$

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MCQ 571 Mark

The vector having initial and terminal points as $(2,5,0)$ and $(-3,7,4)$ respectively is

A
$-\hat{i}+12 \hat{j}+4 \hat{k}$
B
$5 \hat{i}+2 \hat{j}-4 \hat{k}$
✓
$-5 \hat{i}+2 \hat{j}+4 \hat{k}$
D
$\hat{i}+\hat{j}+\hat{k}$

Answer

Correct option: C.

$-5 \hat{i}+2 \hat{j}+4 \hat{k}$

(c) : Let $A(2,5,0)$ and $B(-3,7,4)$
$
\begin{aligned}
\therefore \quad \text { Required vector } & =(-3-2) \hat{i}+(7-5) \hat{j}+(4-0) \hat{k} \\
& =-5 \hat{i}+2 \hat{j}+4 \hat{k}
\end{aligned}
$

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MCQ 581 Mark

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $2 \hat{i}-3 \hat{j}+6 \hat{k}$.

✓
5
B
6
C
7
D
8

Answer

Correct option: A.

(a) : Let $\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}+6 \hat{k}$.
Now, projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$
=\frac{(\hat{i}+3 \hat{j}+7 \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{\sqrt{2^2+(-3)^2+6^2}}=\frac{2-9+42}{\sqrt{4+9+36}}=\frac{35}{7}=5
$

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MCQ 591 Mark

The projection of the vector $\vec{a}=2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k}$ on the vector $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ is

✓
$\frac{10}{\sqrt{6}}$
B
$\frac{10}{\sqrt{3}}$
C
$\frac{5}{\sqrt{6}}$
D
$\frac{5}{\sqrt{3}}$

Answer

Correct option: A.

$\frac{10}{\sqrt{6}}$

(a): We have, $\vec{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$
$
\therefore \quad \vec{a} \cdot \vec{b}=(2 \hat{i}+3 \hat{j}+2 \hat{k}) \cdot(\hat{i}+2 \hat{j}+\hat{k})=2+6+2=10
$
and $|\vec{b}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}$
Hence, projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{10}{\sqrt{6}}$.

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MCQ 601 Mark

If $\vec{b}$ and $\vec{c}$ are any two non-collinear unit vectors and $\vec{a}$ is any vector, then find the value of $(\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} \cdot(\vec{b} \times \vec{c})$.

A
$\vec{a}+\vec{b}+\vec{c}$
B
$\vec{c}$
✓
$\vec{a}$
D
$\vec{b}$

Answer

Correct option: C.

$\vec{a}$

(c) : Let $\vec{b}=\hat{i}$ and $\vec{c}=\hat{j}$ and $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ Now, $\vec{a} \cdot \vec{b}=a_1, \vec{a} \cdot \vec{c}=a_2$ and $\vec{a} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}=\vec{a} \cdot \hat{k}=a_3$
$
\begin{aligned}
\therefore \quad & (\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|}(\vec{b} \times \vec{c}) \\
\quad & =a_1 \vec{b}+a_2 \vec{c}+a_3 \hat{k}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}=\vec{a}
\end{aligned}
$

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MCQ 611 Mark

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is

A
$\frac{3}{2}$
B
3
✓
$\frac{-3}{2}$
D
-3

Answer

Correct option: C.

$\frac{-3}{2}$

(c) : We have $\vec{a}, \vec{b}, \vec{c}$ are unit vectors.
Therefore, $|\vec{a}|=1,|\vec{b}|=1$ and $|\vec{c}|=1$
Also, $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ (given)
$
\begin{array}{l}
\Rightarrow \quad|\vec{a}+\vec{b}+\vec{c}|^2=0 \\
\Rightarrow \quad|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow \quad 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow \quad 3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow \quad(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}
\end{array}
$

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MCQ 621 Mark

The position vector of the point which divides the joining of points $2 \vec{a}-3 \vec{b}$ and $\vec{a}+\vec{b}$ in the ratio $3: 1$ is

A
$\frac{3 \vec{a}-2 \vec{b}}{2}$
B
$\frac{7 \vec{a}-8 \vec{b}}{4}$
C
$\frac{3 \vec{a}}{4}$
✓
$\frac{5 \vec{a}}{4}$

Answer

Correct option: D.

$\frac{5 \vec{a}}{4}$

(d) : Given points are $2 \vec{a}-3 \vec{b}$ and $\vec{a}+\vec{b}$.
$
\begin{array}{l}
\text { Given ratio }=3: 1 \\
\therefore \quad \text { Required vector }=\frac{(2 \vec{a}-3 \vec{b}) \times 1+(\vec{a}+\vec{b}) \times 3}{3+1} \\
=\frac{2 \vec{a}-3 \vec{b}+3 \vec{a}+3 \vec{b}}{4}=\frac{5}{4} \vec{a} \\
\end{array}
$

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Maths M.C.Q (1 Marks)