M.C.Q (1 Marks)

Question 1011 Mark

A set of vectors taken in a given order gives a closed polygon.Then the resultant of these vectors is:

scalar quantity
pseudo vector
unit vector
null vector

Answer

null vector

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Question 1021 Mark

If $\vec{a}\ \text{and}\ \vec{b}$ are two collinear vectors, then which of the following are incorrect:

$\vec{b}=\lambda\vec{a},\ \text{for some scalar}\ \lambda$
$\vec{a}=\pm\vec{b}$
The respective components of $\vec{a}\ \text{and}\ \vec{b}$ are proportional.
Both the vectors $\vec{a}\ \text{and}\ \vec{b}$ have same direction, but different magnitudes.

Answer

If $\vec{a}\ \text{and}\ \vec{b}$ are two collinear vectors, then they are parallel. Therefore, we have: $\vec{b}=\lambda\vec{a}\ (\text{For some scalar}\ \lambda)$ $\text{If}\ \lambda=\pm1,\ \text{then}\ \vec{a}=\pm\vec{b}$ $\text{If}\ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\ \text{and}\ \vec{b}$ $=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}, \text{then}\ \vec{b}=\lambda\vec{a}.$ $\Rightarrow{b_1}\hat{i}+b_2\hat{j}+b_3\hat{k}=\lambda\big({a_1}\hat{i}+a_2\hat{j}+a_3\hat{k}\big)$ $\Rightarrow{b_1}\hat{i}+b_2\hat{j}+b_3\hat{k}=\big(\lambda{a_1}\big)\hat{i}+\big(\lambda{a_2}\big)\hat{j}+\big(\lambda{a_3}\big)\hat{k}$ $\Rightarrow{b_1}=\lambda{a_1,}\ b_2=\lambda{a_2,}\ b_3=\lambda{a_3}$ $\Rightarrow\frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}=\lambda$
Thus, the respective components of $\vec{a}\ \text{and}\ \vec{b}$ are proportional. However, vectors $\vec{a}\ \text{and}\ \vec{b}$ can have different directions. Hence, the statement given in D is incorrect. The correct answer is D.

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Question 1031 Mark

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

0
-1
1
3

Answer

$\hat{\text{i}}\cdot\Big(\hat{\text{j}}\times\hat{\text{k}}\Big)+\hat{\text{j}}\cdot\Big(\hat{\text{i}}\times\hat{\text{k}}\Big)+\hat{\text{k}}\cdot\Big(\hat{\text{i}}\times\hat{\text{j}}\Big)$
$=\hat{\text{i}}\cdot\hat{\text{i}}+\hat{\text{j}}\cdot\Big(-\hat{\text{j}}\Big)+\hat{\text{k}}\cdot\hat{\text{k}}$
$=1-\hat{\text{j}}\cdot\hat{\text{j}}+1$
=1-1+1
=1
The correct answer is C.

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Question 1041 Mark

Choose the correct answer
If $\theta$ is the angle between any two vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}, \text{then}\ |\vec{\text{a}}\cdot\vec{\text{b}}|=|\vec{\text{a}}\times\vec{\text{b}}|\ \text{when}\ \theta$ is equal to

0
$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\pi$

Answer

Let $\theta$ be the angle between two vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}.$
Then, without loss of generality, $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ are non-zero vectors, so that $\big|\vec{\text{a}}\big|\ \text{and}\ \Big|\vec{\text{b}}\Big|$ are position.
$\Big|\vec{\text{a}}\cdot\vec{\text{b}}\Big|=\Big|\vec{\text{a}}\times\vec{\text{b}}\Big|$
$\Rightarrow\big|\vec{\text{a}}\big|\Big|\vec{\text{b}}\Big|\cos\theta=\big|\vec{\text{a}}\big|\Big|\vec{\text{b}}\Big|\sin\theta$
$\Rightarrow\cos\theta=\sin\theta\ \ \ \Big[\big|\vec{\text{a}}\big|\ \text{and}\ \Big|\vec{\text{b}}\Big|\Big]\ \text{are positive}$
$\Rightarrow\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
Hence, $\Big|\vec{\text{a}}.\vec{\text{b}}\Big|=\Big|\vec{\text{a}}\times\vec{\text{b}}\Big|$ when $\theta$ is equal to $\frac{\pi}{4}.$
The correct answer is B.

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Question 1051 Mark

If $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\text{c}}=5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}},$ then the volume of the parallelopiped with contermious edges $\vec{\text{a}}+\vec{\text{b}},\vec{\text{b}}+\vec{\text{c}},\vec{\text{c}}+\vec{\text{a}}$ is:

2
1
-1
0

Answer

We have
$\vec{\text{a}}+\vec{\text{b}}=\big(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\big)+\big(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)=5\hat{\text{i}}-7\hat{\text{j}}+10\hat{\text{k}}$
$\vec{\text{b}}+\vec{\text{c}}=\big(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)\big(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)=8\hat{\text{i}}-7\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{c}}+\vec{\text{a}}=\big(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+\big(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\big)=7\hat{\text{i}}-6\hat{\text{j}}+3\hat{\text{k}}$
We know that the volume of parallelopiped whose three adjacent adges are $\vec{\text{a}}+\vec{\text{b}},\vec{\text{b}}+\vec{\text{c}}$ and $\vec{\text{c}}+\vec{\text{a}}$ is equal to:
We have
$\big[\vec{\text{a}}+\vec{\text{b}}\vec{\text{b}}+\vec{\text{c}}\vec{\text{c}}+\vec{\text{a}}\big]=\begin{vmatrix}5&-7&10\\8&-7&3\\7&-6&3\end{vmatrix}$
$=5(-21+18)+7(24-21)+10(-48+49)$
$=(5\times-3)+(7\times3)+(10\times1)$
$=16$
$\therefore$ volume of parallelopiped $=\big|16\big|=16$
Disclaimer: None of the given option is correct.

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Question 1061 Mark

Let the vectors $\vec{a}\ \text{and}\ \vec{b}$ be such that $|\vec{a}|=3\ \text{and}\ \big|\vec{b}\big|=\frac{\sqrt{2}}{3},\ \text{then}\ \vec{a}\times\vec{b}$ is a unit vector, if the angle between $\vec{a}\ \text{and}\ \vec{b}\ \text{is}$

$\pi/6$
$\pi/4$
$\pi/3$
$\pi/2$

Answer

$\text{Given:}\ \ \big|\vec{a}\big|=3,\big|\vec{b}\big|=\frac{\sqrt{2}}{3}\ \text{and}\ \vec{a}\times\vec{b}$ is a unit vector.
$\Rightarrow\ \ \big|\vec{a}\times\vec{b}\big|=1\ $ $\Rightarrow\ \big|\vec{a}\big|.\Big|\vec{b}\Big|\ \text{sin}\ \theta=1,\ \text{where}\ \theta\ \text{is the angle between}\ \vec{a}\ \text{and}\ \vec{b}.$
$\Rightarrow\ \ 3\bigg(\frac{\sqrt{2}}{3}\bigg)\ \text{sin}\ \theta=1$ $\Rightarrow\ \sqrt{2}\ \text{sin}\ \theta=1\ \Rightarrow\ \ \text{sin}\ \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow\ \text{sin}\ \theta=\text{sin}\frac{\pi}{{4}}\ \ \Rightarrow\ \theta=\frac{\pi}{4}$
Therefore, option (B) is correct.

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Question 1071 Mark

If $\vec{a}$ is a nonzero vector of magnitude 'a' and $\lambda$ a nonzero scalar, then $\lambda\ \vec{a}$ is unit vector if

$\lambda=1$
$\lambda=-1$
$a=\big|\lambda\big|$
$a=1/\big|\lambda\big|$

Answer

Given: $ \vec{a}$ is a non-zero vector of magnitude a $ \Rightarrow\ \ \ |\vec{a}|=1$
Also given $\lambda\neq0\ \text{and}\ \lambda\vec{a}$ is a unit vector.
$\Rightarrow\ \ |\lambda\vec{a}|=1\ \Rightarrow\ \ |\lambda|\big|\vec{a}\big|=1$
$\Rightarrow\ \ \ \ \ \ |\lambda|a=1\ \ \Rightarrow\ \ a=\frac{1}{|\lambda|}$
Therefore, option (D) is correct.

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Question 1081 Mark

Point (4, 0) lies on ________?

$\vec{\text{XO}}$
$\vec{\text{YO}}$
$\vec{\text{OX}}$
$\vec{\text{OY}}$

Answer

$\vec{\text{OX}}$

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MCQ 1091 Mark

Two vectors each of magnitudes $1$ unit are inclined at $60^\circ$ to each other. The difference of the vectors has a magnitude $........$?

A
$0$ units
B
$1$ units
✓
$2$ units
D
$3$ units

Answer

Correct option: C.

$2$ units

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Question 1101 Mark

If $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$ and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}.\vec{\text{a}}\neq0,$ then:

$\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{b}}=\vec{0}$
$\vec{\text{b}}+\vec{\text{c}}=\vec{0}$
$\text{None of these}$

Answer

$\vec{\text{b}}=\vec{\text{c}}$

Solution:
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}-\vec{\text{a}}.\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}.\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\big(\vec{\text{b}}-\vec{\text{c}}\big)$
$|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta\dots(1)$
and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Then, $|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0\dots(2)$
Here, it is given that $\vec{\text{a}}\neq0$
Therefore, for eq. (1) and eq. (2) to be 0
We have,
$\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0$
For $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0,$ one of $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|$ or $\cos\theta$ must be 0
Case 1 :
Let $\cos\theta=0$
$\Rightarrow\theta=90^\circ$
$\Rightarrow\sin\theta=1$
& if $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0$ and $\sin\theta=1$
Then $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Case 2 :
Let $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Hence, $\vec{\text{b}}=\vec{\text{c}}$

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Question 1111 Mark

Can two different vectors have the same magnitude:

Yes
No
Cannot be determined
None of the above

Answer

Solution:
Two vectors can have the same magnitude.
Magnitude of vector i - 2j + k is equal to magnitude of vector 2i + j - k.

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Question 1121 Mark

If $\theta$ is the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ then $\vec{\text{a}}.\vec{\text{b}}\geq0$ only when:

$0<\theta\frac{\pi}{2}$
$0\leq\theta\leq\frac{\pi}{2}$
$0<\theta<\pi$
$0\leq\theta\leq\pi$

Answer

$0\leq\theta\leq\frac{\pi}{2}$

Solution:
$\vec{\text{a}}.\vec{\text{b}}\geq0$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\geq0$
$\Rightarrow\cos\theta\geq0$
$\Rightarrow0\leq\theta\leq\frac{\pi}{2}$

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Question 1131 Mark

If $\vec{\text{a}}$ is any vector, then $\big(\vec{\text{a}}\times\hat{\text{i}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{j}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{k}}\big)^2=$

$\vec{\text{a}}^2$
$2\vec{\text{a}}^2$
$3\vec{\text{a}}^2$
$4\vec{\text{a}}^2$

Answer

$2\vec{\text{a}}^2$

Solution:
Let $\vec{\text{a}}={\text{a}}_1\hat{\text{i}}+{\text{a}}_2\hat{\text{j}}+{\text{a}}_3\hat{\text{k}}$
$\vec{\text{a}}\times\hat{\text{i}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\1&0&0 \end{vmatrix}$
$=\text{a}_3\hat{\text{j}}-\text{a}_2\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\hat{\text{i}}\big)^2=\big(\text{a}_3\hat{\text{j}}-\text{a}_2\hat{\text{k}}\big)^2$
$={\text{a}_3}^2|\hat{\text{j}}|^2+{\text{a}_2}^2|\hat{\text{k}}|^2-2\text{a}_3\text{a}_2\big(\hat{\text{j}}.\hat{\text{k}}\big)$
$={\text{a}_3}^2+{\text{a}_2}^2$ $\big(\because\hat{\text{j}}.\hat{\text{k}}=0\dots(1)\big)$
$\therefore\vec{\text{a}}\times\hat{\text{j}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\0&1&0 \end{vmatrix}$
$=-\text{a}_3\hat{\text{i}}+\text{a}_1\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\hat{\text{j}}\big)^2=\big(-\text{a}_3\hat{\text{i}}+\text{a}_1\hat{\text{k}}\big)^2$
$={\text{a}_3}^2|\hat{\text{i}}|^2+{\text{a}_1}^2|\hat{\text{k}}|^2-2\text{a}_3\text{a}_2\big(\hat{\text{i}}.\hat{\text{k}}\big)$
$={\text{a}_3}^2+{\text{a}_1}^2$ $(\because\hat{\text{i}}.\hat{\text{k}}=0)\dots(2)$
$\therefore\vec{\text{a}}\times\hat{\text{k}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\0&0&1 \end{vmatrix}$
$=\text{a}_2\hat{\text{i}}-\text{a}_1\hat{\text{j}}$
$\Rightarrow\big(\vec{\text{a}}\times\hat{\text{k}}\big)^2=\big(\text{a}_2\hat{\text{i}}-\text{a}_1\hat{\text{j}}\big)^2$
$={\text{a}_2}^2|\hat{\text{i}}|^2+{\text{a}_1}^2|\hat{\text{j}}|^2-2\text{a}_1\text{a}_2\big(\hat{\text{i}}.\hat{\text{j}}\big)$
$={\text{a}_2}^2+{\text{a}_1}^2$ $(\because\hat{\text{i}}.\hat{\text{j}}=0)\dots(3)$
Adding (1), (2) and (3), we get
$\big(\vec{\text{a}}\times\hat{\text{i}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{j}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{k}}\big)^2={\text{a}_3}^2+{\text{a}_2}^2+{\text{a}_3}^2+{\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_1}^2$
$=2\big({\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_3}^2\big)$
$=2\vec{\text{a}}^2$ $\big(\because|\vec{\text{a}}|=\sqrt{{\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_3}^2}\big)$

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Question 1141 Mark

In a regular hexagon ABCDEF, $\overrightarrow{\text{AB}}=\vec{\text{a}},\ \overrightarrow{\text{BC}}=\vec{\text{b}}$ and $\overrightarrow{\text{CD}}=\vec{\text{c}}$. Then, $\overrightarrow{\text{AE}}=$

$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
$2\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
$\vec{\text{b}}+\vec{\text{c}}$
$\vec{\text{a}}+2\vec{\text{b}}+2\vec{\text{c}}$

Answer

$\vec{\text{b}}+\vec{\text{c}}$

Solution:
Given a regular hexagon ABCDEF, $\overrightarrow{\text{AB}}=\vec{\text{a}},\ \overrightarrow{\text{BC}}=\vec{\text{b}}$ and $\overrightarrow{\text{CD}}=\vec{\text{c}}$. Then,
In $\triangle{\text{ABC}}$, we have
$\overrightarrow{\text{AC}}=\vec{\text{a}}+\vec{\text{b}}$
In $\triangle{\text{ACD}}$, we have
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{AC}}+\vec{\text{c}}$
$\Rightarrow\overrightarrow{\text{AD}}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
Again, in $\triangle{\text{ADE}}$, we have
$\overrightarrow{\text{AE}}=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}$
$\Rightarrow\overrightarrow{\text{AE}}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AE}}=\vec{\text{b}}+\vec{\text{c}}$
Hence option (c).

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Question 1151 Mark

If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two collinear vectors, then which of the follwoing are incorrect?

$\vec{\text{b}}=\lambda\vec{\text{a}}$ for some scalar $\lambda$
$\vec{\text{a}}=\pm\vec{\text{b}}$
The respective components of $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are proportional.
Both the vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ have the same direction but different magnitudes.

Answer

Both the vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ have the same direction but different magnitudes.

Solution:
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are collinear vectors, then they are parallel. Therefore, we have $\vec{\text{b}}=\lambda\vec{\text{a}}$, for some scalar $\lambda$.
If $\lambda=\pm1\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
If $\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$. Then,
$\vec{\text{b}}=\lambda\vec{\text{a}}$
$\Rightarrow\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}=\lambda\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)$
$\Rightarrow\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}=(\lambda\text{a}_1)\hat{\text{i}}+(\lambda\text{a}_2)\hat{\text{j}}+(\lambda\text{a}_3)\hat{\text{k}}$
$\Rightarrow\ \text{b}_1=\lambda\text{a}_1,\ \text{b}_2=\lambda\text{a}_2,\ \text{b}_3=\lambda\text{a}_3$
$\Rightarrow\ \frac{\text{b}_1}{\text{a}_1}=\frac{\text{b}_2}{\text{a}_2}=\frac{\text{b}_3}{\text{a}_3}=\lambda$
Thus, the respective components of $\vec{\text{a}}\text{ and }\vec{\text{b}}$ can have different directions. Hence, the statement given in (d) is incorrect.

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Question 1161 Mark

The curve y-x:

A vertical tangent (parallel to y-axis)
A horizontal tangent (parallel to x-axis)
An oblique tangent
No tangent

Answer

A horizontal tangent (parallel to x-axis)

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Question 1171 Mark

If three points A, B and C have position vectors $\hat{\text{i}}+\text{x}\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$ respectively are collinear, then (x, y) =

(2, -3)
(-2, 3)
(-2, -3)
(2, 3)

Answer

(2, -3)

Solution:
Given position vector of A, B and C are $\hat{\text{i}}+\text{x}\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$. Then,
$\overrightarrow{\text{AB}}=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-\text{x}\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+(4-\text{x})\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\text{y}\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=(\text{y}-3)\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
Since, the given vectors are collinear.
$\therefore\ \overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$\Rightarrow\ 2\hat{\text{i}}+(4-\text{x})\hat{\text{j}}+4\hat{\text{k}}=\lambda(\text{y}-3)\hat{\text{i}}-6\lambda\hat{\text{j}}-12\lambda\hat{\text{k}}$
$\Rightarrow\ 2=\lambda(\text{y}-3),\ (4-\text{x})=-6\lambda,\ 4=-12\lambda$
$\Rightarrow\ 2=\lambda(\text{y}-3),\ (4-\text{x})=-6\lambda,\ \lambda=-\frac{1}3$
$\Rightarrow\ 2=-\frac{1}3(\text{y}-3),\ (4-\text{x})=-6\times\Big(-\frac{1}3\Big)$
$\Rightarrow\ -6=\text{y}-3,\ 4-\text{x}=2$
$\Rightarrow\ \text{y}=-3,\ \text{x}=2$

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Question 1181 Mark

A point from a vector starts is called and where it ends is called its:

Terminal point, endpoint.
Initial point, terminal point
Origin, endpoint
Initial point, endpoint

Answer

Initial point, terminal point

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Question 1191 Mark

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

1
0
2
-1
-2

Answer

Solution:
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{j}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).\hat{\text{j}}}{|\hat{\text{j}}|}$
$=\frac{0+1+0}{1}$
$=1$

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Question 1201 Mark

The position vectors of the points A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ respectively. These points,

Form an isosceles triangle.
Form a right triangle.
Are collinear.
Form a scalene triangle.

Answer

Form an isosceles triangle.

Solution:
Given: Position vectors of A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$. Then,
$\overrightarrow{\text{AB}}=\big(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)-\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\big(\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)-\big(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$
$=-2\hat{\text{i}}+6\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{CA}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)-\big(\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)$
$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{1^2+(-3)^2+2^2}$
$=\sqrt{1+9+4}$
$=\sqrt{14}$
$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{1^2+(-3)^2+2^2}$
$=\sqrt{1+9+4}$
$=\sqrt{14}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(-2)^2+6^2+(-4)^2}$
$=\sqrt{4+36+16}$
$=\sqrt{56}$
$\therefore\Big|\overrightarrow{\text{AB}}\Big|=\Big|\overrightarrow{\text{CA}}\Big|$
Hence, the triangle is isosceles as two of its sides are equal.

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Question 1211 Mark

If $\mid\text{a}\mid=5,\mid\text{b}\mid=13$ and $\mid\text{a}\times{\text{b}}\mid=25$ find a.b:

$\underline{+}10$
$\underline{+}40$
$\underline{+}60$
$\underline{+}25$

Answer

$\underline{+}60$

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Question 1221 Mark

If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, then $\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big[\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{a}}+\vec{\text{c}}\big)\big]$ equals:

$0$
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

Answer

$-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

Solution:
$\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big[\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{a}}+\vec{\text{c}}\big)\big]$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=0+0+\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]+\big[\vec{\text{b}}\vec{\text{a}}\vec{\text{c}}\big]+0+0+0+\big[\vec{\text{c}}\vec{\text{b}}\vec{\text{a}}\big]+0$
$=-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

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Question 1231 Mark

If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, then which of the following values of $\vec{\text{a}}.\vec{\text{b}}$ is not possible?

$\sqrt{3}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{-1}{2}$

Answer

$\sqrt{3}$

Solution:
It is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors.
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1$
Now,
$\vec{\text{a}}.\vec{\text{b}}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$=(1)(2)\cos\theta$
$=\cos\theta$
The range of $\cos\theta$ is [-1,1].
$\therefore\sqrt{3}$ is not a possible value of $\cos\theta$ as it is greater than 1.

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Question 1241 Mark

The vectors $2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ are perpendicular if:

a = 2, b = 3, c = -4
a = 4, b = 4, c = 5
a = 4, b = 4, c = -5
a = -4, b = 4, c = -5

Answer

a = 4, b = 4, c = 5

Solution:
It is given that vectors $2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ are perpendicular.
So, their dot product is zero.
$\Rightarrow2\text{a}+3\text{b}-4\text{c}=0$
$(\text{b})\text{a}=4;\text{b}=4;\text{c}=5$
$\Rightarrow2(4)+3(4)-4(5)=0$
$8+12-20=0$
$0=0,$ which is true.

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Question 1251 Mark

The resultant of two concurrent forces $\vec{\text{nOP}}$ and $\vec{\text{mOQ}}$ is$(\text{m+n})\vec{\text{OR.}}$ Then R divides PQ in the ratio:

m : n
n : m
1 : n
m : 1

Answer

m : n

Solution:
Applying Section Formula
$\text{R}=\frac{\text{KQ}+\text{P}}{\text{K}+1}$
$(\text{K+1})\text{R = KQ + P}$
$\text{K+1}=\frac{\text{m+n}}{\text{n}}$
$\text{K}=\frac{\text{m}}{\text{n}}$

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Question 1261 Mark

The unit vector in the direction of $\overrightarrow{\text{a}}$ is:

$\frac{\vec{\text{a}}}{\mid\vec{\text{a}\mid}}$
$\vec{\text{a}}\mid\vec{\text{a}}\mid$
$\text{a}^{2}$
$\hat{\text{i}}$

Answer

$\frac{\vec{\text{a}}}{\mid\vec{\text{a}\mid}}$

Solution:
Consider the given vector $\vec{\text{a}}$
Unit vector $\hat{\text{a}}$

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Question 1271 Mark

$\overrightarrow{\text{r}} = \overrightarrow{\text{x}}{\hat{\text{i}}}+ \overrightarrow{\text{y}}{\hat{\text{j}}}$ is the equation of:

Yoz plane
A straight line joining the points ${\hat{\text{i}}}$ and ${\hat{\text{j}}}$
Zox plane
Xoy plane

Answer

Xoy plane

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Question 1281 Mark

Time period is a:

Vector quantity
Scalar quantity
Neither scalar nor vector
None of these

Answer

Scalar quantity

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Question 1291 Mark

If $\theta$ is the angle between the vectors $2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$ and $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},$ then $\sin\theta=$

$\frac{2}{3}$
$\frac{2}{\sqrt{7}}$
$\frac{\sqrt{2}}{7}$
$\sqrt{\frac{2}{7}}$

Answer

$\frac{2}{\sqrt{7}}$

Solution:
Let:
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-2)^2+4^2}$
$=\sqrt{4+4+16}$
$=\sqrt{24}$
$=2\sqrt{6}$
$\Big|\vec{\text{b}}\big|=\sqrt{3^2+1^2+2^2}$
$=\sqrt{9+1+4}$
$=\sqrt{14}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-2&4\\3&1&2 \end{vmatrix}$
$=-8\hat{\text{i}}+8\hat{\text{j}}+8\hat{\text{k}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{64+64+64}$
$=\sqrt{192}$
$=8\sqrt{3}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow8\sqrt{3}=(2\sqrt{6})(\sqrt{14})\sin\theta$
$\Rightarrow\sin\theta=\frac{8\sqrt{3}}{4\sqrt{21}}$
$=\frac{2}{\sqrt{7}}$
$\Rightarrow\theta=\sin^{-1}\Big(\frac{2}{\sqrt{7}}\Big)$

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Question 1301 Mark

Which of the given qualities is a vector:

Speed
Time
Weight
Volume

Answer

Speed

Solution:
Speed is a vector quantity as it has both magnitude and direction. Time, weight, volume have only magnitude and no direction. they all are scalar quantity.

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Question 1311 Mark

If vectors $(\text{x}-2)\ \vec{\text{a}}+\vec{\text{b}}$ and $(2\text{x}+1)\ \vec{\text{a}}-\vec{\text{b}}$ are parallel then x:

$\frac{1}{3}$
$3$
$-3$
$\frac{-1}{3}$

Answer

$\frac{1}{3}$

Solution:
As vectors (x - 2) a + b and (2x + 1) a - b are parallel.
$\frac{\text{x}-2}{2\text{x}+1}=-1$
$\Rightarrow\text{x - 2}=-2\text{x}-1$
$\therefore\text{x}=\frac{1}{3}$

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Question 1321 Mark

Choose the correct answer from the given four options.
Projection vector of $\vec{\text{a}}$ on $\vec{\text{b}}$ is:

$\bigg(\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{b}}|^2}\bigg)\vec{\text{b}}$
$\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{b}}|}$
$\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{a}}|}$
$\bigg(\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{a}}|^2}\bigg)\vec{\text{b}}$

Answer

$\bigg(\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{b}}|^2}\bigg)\vec{\text{b}}$

Solution:
Projection vector of $\vec{\text{a}}$ on $\vec{\text{b}}$ is given by $\vec{\text{a}}\cdot\frac{\vec{\text{b}}}{|\vec{\text{b}}|}\vec{\text{b}}=\bigg(\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{b}}|^2}\bigg)\vec{\text{b}}$

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MCQ 1331 Mark

Choose the correct answer from the given four options. Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is:

A
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{3}$
✓
$\frac{4}{7}$

Answer

Correct option: D.

$\frac{4}{7}$

Here, $S = \{(B, B, B), (G, G, G), (B, G, G), (G, B, G), (G, G, B), (G. B, B), (B, G, B), (B, B. G)\}$
$E_1 =$ Event that a family has atleast one girl, then
$E_1 = \{(G, B, B), (B, G, B), (B, B. G), (G, G, B), (B, G, G), (G. B, G), (G, G, G)\}$
$E_2 =$ Event that the eldest child is a girl, then
$E_2 = \{(G, B, B), (G, G, B), (G, B, G) (G, G, G)\}$
$\therefore\text{E}_1\cup\text{E}_2=\left\{(\text{G},\text{B},\text{B}),(\text{G},\text{G},\text{B}),(\text{G},\text{B},\text{G}),(\text{G},\text{G},\text{G})\right\}$
$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_1)}$
$=\frac{\frac{4}{8}}{\frac{7}{8}}$
$=\frac{4}{7}$

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Question 1341 Mark

If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors,then the greatest value of $\sqrt{3}\big|\vec{\text{a}}+\vec{\text{b}}\big|+\big|\vec{\text{a}}-\vec{\text{b}}\big|$ is:

$2$
$2\sqrt{2}$
$4$
$\text{None of these}$

Answer

Solution:
We have
$\sqrt{3}\big|\vec{\text{a}}+\vec{\text{b}}\big|+\big|\vec{\text{a}}-\vec{\text{b}}\big|$
$=\sqrt{3}\times\sqrt{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta}+\sqrt{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta}$
$=\sqrt{3}\times\sqrt{1^2+1^2+2\times1\times1\cos\theta}+\sqrt{1^2+1^2-2\times1\times1\cos\theta}$ (As $\vec{\text{a}}$ and $\vec{\text{b}}$ unit vectors)
$=\sqrt{3}\times\sqrt{2+2\cos\theta}+\sqrt{2-2\cos\theta}$
$=\sqrt{3}\times\sqrt{2(1+\cos\theta)}+\sqrt{2(1-\cos\theta)}$
$=\sqrt{3}\times\sqrt{2\times2\cos^2\frac{\theta}{2}}+\sqrt{2\times2\sin^2\frac{\theta}{2}}$
$=2\sqrt{3}\cos\frac{\theta}{2}+2\sin\frac{\theta}{2}$
$=2\big(\sqrt{3}\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\big)$
$=2\times2\big(\frac{\sqrt{3}}{2}\cos\frac{\theta}{2}+\frac{1}{2}\sin\frac{\theta}{2}\big)$
$=2\times2\big(\sin\frac{\pi}{3}\cos\frac{\theta}{2}+\cos\frac{\pi}{3}\sin\frac{\theta}{2}\big)$
$=4\sin\big(\frac{\pi}{3}+\frac{\theta}{2}\big)$
Now, maximum value of $\sin\text{a}=1$
⇒ Maximum value of $\sin\big(\frac{\pi}{3}+\frac{\theta}{2}\big)=1$
⇒ Maximum value of $4\sin\big(\frac{\pi}{3}+\frac{\theta}{2}\big)=4$
$\therefore$ Maximum velue of $\sqrt{3}\big|\vec{\text{a}}+\vec{\text{b}}\big|+\big|\vec{\text{a}}-\vec{\text{b}}\big|=4$

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Question 1351 Mark

If the vectors $\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}$ and $\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}$ are perpendicular, then the locus of (x,y) is:

A circle.
An ellipse.
A hyperbola.
None of these.

Answer

an ellipse

Solution:
Let, $\vec{\text{a}}=\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}$
It is given that the vectors are perpendicular. so, their dot product is zero.
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\big(\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}\big).\big(\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}\big)=0$
$\Rightarrow1-4\text{x}^2-9\text{y}^2=0$
$\Rightarrow4\text{x}^2+9\text{y}^2=1$
Dividing both sides by 36, we get
$\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=1$
This is an ellipse.

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Question 1361 Mark

If $\mid\text{a}\times\text{b}\mid=4$ and $\mid\text{a.b}\mid=2$ then $\mid{\text{a}}\mid^2\mid{\text{b}}\mid^2$ is equal to:

Answer

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Question 1371 Mark

Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be three unit vectors, such that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$ and $\vec{\text{a}}$ is perpendicular to $\vec{\text{b}}.$ If $\vec{\text{c}}$ makes angle $\alpha$ and $\beta$ with $\vec{\text{a}}$ and $\vec{\text{b}}$ respectively, then $\cos\alpha+\cos\beta=$

$-\frac{3}{2}$
$\frac{3}{2}$
$1$
$-1$

Answer

$-1$

Solution:
Given that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=1$ and $\vec{\text{c}}=1.$
Since $\vec{\text{a}}$ and $\vec{\text{b}}$ are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=0$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}=1$
$\Rightarrow1+1+1+2(0)+2|\vec{\text{a}}|\big|\vec{\text{b}\big|}\cos\beta+2|\vec{\text{c}}||\vec{\text{a}}|\cos\alpha=1$
$\Rightarrow3+2(\cos\alpha+\cos\beta)=1$
$\Rightarrow2(\cos\alpha+\cos\beta)=-2$
$\Rightarrow\cos\alpha+\cos\beta=-1$

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Question 1381 Mark

$\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big\{\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{a}}-\vec{\text{b}}-\vec{\text{c}}\big)\big\}$ is equal to:

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
$2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
$3\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
$0$

Answer

$3\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$

Solution:
$\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big\{\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{a}}-\vec{\text{b}}-\vec{\text{c}}\big)\big\}$
$=\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{a}}-\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big(-\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big(-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big[\text{a}\text{b}\text{c}\big]+2\big[\text{a}\text{b}\text{c}\big]$
$=3\big[\text{a}\text{b}\text{c}\big]$

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Question 1391 Mark

If $\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big],$ then $\lambda+\mu=$

6
-6
10
8

Answer

Solution:
We have
$\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[\big(2\vec{\text{a}}+4\vec{\text{b}}\big]\times\vec{\text{c}}\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$ (By definition of scalar triple product)
$\Rightarrow\big[\big(2\vec{\text{a}}\times\vec{\text{c}}\big)+\big(4\vec{\text{b}}\times\vec{\text{c}}\big)\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big(2\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{d}}+\big(4\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[2\vec{\text{a}}\vec{\text{ c }}\vec{\text{d}}\big]+\big[4\vec{\text{b}}\vec{\text{ c }}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{ c }}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{ c }}\vec{\text{d}}\big]$
$\Rightarrow2\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+4\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$ $\big(\therefore\big[\lambda\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ for any scaler $\lambda\big)$
Comparing both sides, we get
$\lambda=2$
$\mu=4$
$\therefore\lambda+\mu=2+4=6$

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Question 1401 Mark

A unit vector perpendicular to both $\hat{\text{i}}+\hat{\text{j}}$ and $\hat{\text{j}}+\hat{\text{k}}$ is:

$\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\frac{1}{\sqrt{3}}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$

Answer

$\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$

Solution:
Let:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&0\\0&1&1 \end{vmatrix}$
$=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{1+1+1}$
$=\sqrt{3}$
Unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}=\frac{\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}}{\sqrt{3}}$
Disclaimer: The answer given for this question in the textbook is incorrect.

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Question 1411 Mark

Choose the correct answer from the given four options.
The value of $\lambda$ for which the vectors $3\hat{\text{i}}-6\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-4\hat{\text{j}}+\lambda\hat{\text{k}}$ are parallel, is:

$\frac{2}{3}$
$\frac{3}{2}$
$\frac{5}{2}$
$\frac{2}{5}$

Answer

$\frac{2}{3}$

Solution:
As the vectors $3\hat{\text{i}}-6\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-4\hat{\text{j}}+\lambda\hat{\text{k}}$ are parallel
$\therefore\frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda}$
$\Rightarrow\lambda=\frac{2}{3}$

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Question 1421 Mark

If a, b, c are position vectors of the vertices of a $\Delta\text{ABC}$ then $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=$

Answer

Solution:
If we join head to tail all the vectors, then we end up at the initial point where we started, that is vertice A. the net sum is 0.

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Question 1431 Mark

The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vector is:

$\sqrt{3}$
$1-\sqrt{3}$
$1+\sqrt{3}$
$-\sqrt{3}$

Answer

$\sqrt{3}$

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Question 1441 Mark

If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=4,\big|\vec{\text{a}}.\vec{\text{b}}\big|=2,$ then $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=$

Answer

Solution:
We know
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|62=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\dots(1)$
$\big|\vec{\text{a}}.\vec{\text{b}}\big|=2$ (Given)
$\Rightarrow\big|\vec{\text{a}}.\vec{\text{b}}\big|^2=\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
From (1), we get
$(2)^2+(4)^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=20$

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Question 1451 Mark

For non-zero vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good, if:

$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=0$
$\vec{\text{a}}.\vec{\text{b}}=0=\vec{\text{c}}.\vec{\text{a}}$
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
$\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$

Answer

$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$

Solution:
we have
$\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|\big|\cos\theta\big|$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|$ (If $\theta=0^\circ\text{or}180^\circ,$ i.e. vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{c}}$ are parallel)
$=\big|\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\alpha\big)\big|\big|\vec{\text{c}}\big|$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ (If $\alpha=90^\circ,$ i. e. vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular)
$\therefore\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ (If vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other)
Thus, the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good if $\vec{\text{a}}.\vec{\text{b}}=0,\vec{\text{b}}.\vec{\text{c}}=0$ and $\vec{\text{c}}.\vec{\text{a}}=0.$

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Question 1461 Mark

Line passing through (3, 4, 5) and (4, 5, 6) has direction ratios:

$1,1,1$
$\sqrt{3},\sqrt{3},\sqrt{3}$
$\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
$7,9,11$

Answer

$1,1,1$

Solution:
Given points (3, 4, 5) and (4, 5, 6) The drs are given as (4 - 3, 5 - 4, 6 - 5) = (1, 1, 1)

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Question 1471 Mark

If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$ is equal to:

0
2
1
None of these

Answer

Solution:
We have
$\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
$\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}+\frac{\big[\vec{\text{b}}\vec{\text{a}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}$ (By definition of scalar tiple product)
$=\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}+\frac{-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}$ (Change in cyclic order of vectors changes the sign of the scalar triple product)
$=1-1$
$=0$

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Question 1481 Mark

A zero vector has:

Any direction
No direction
Many direction
None of these

Answer

Any direction

Solution:

Zero vector, is a vector of length 0, and thus has all components equal to zero. It is the additive identity of the additive group of vectors.
Thus, it has zero magnitude and arbitrary direction.

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Question 1491 Mark

$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)=$

$0$
$-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

Answer

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

Solution:
We have
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}.\big[\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{a}}+\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{b}}+\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{c}}\big]$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}+0+\vec{\text{c}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+0$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}-\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)+\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$
$=0+0+0+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

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Question 1501 Mark

Choose the correct answer from the given four options.
The angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ is:

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{5\pi}{2}$

Answer

$\frac{\pi}{3}$

Solution:
Here, $|\vec{\text{a}}|=\sqrt{3},|\vec{\text{b}}|=4$ and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ [given]
We know that, $\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\Rightarrow2\sqrt{3}=\sqrt{3}.4.\cos\theta$
$\Rightarrow\cos\theta=\frac{2\sqrt{3}}{4\sqrt{3}}=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}$

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MATHS M.C.Q (1 Marks)