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

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M.C.Q (1 Marks)

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Question 11 Mark

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

$0$
$1$
$\frac{-2}{3}$
$\frac{-3}{2}$

Answer

$\frac{-2}{3}$

Solution:
Since, two non zero vector $\vec{\text{a}}\ \&\ \vec{\text{b}}$ are i.e.,
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$
$\vec{\text{a}}.\vec{\text{b}}=0$
$(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}).(2\hat{\text{i}}+\lambda\hat{\text{k}})=0$
$2+3\lambda=0$
$-2=3\lambda$
$\lambda=\frac{-2}{3}$

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Question 21 Mark

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

$0$
$1$
$\frac{-2}{3}$
$\frac{-3}{2}$

Answer

$\frac{-2}{3}$

Solution:
Since, two non zero vector $\vec{\text{a}}\ \&\ \vec{\text{b}}$ are i.e.,
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$
$\vec{\text{a}}.\vec{\text{b}}=0$
$(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}).(2\hat{\text{i}}+\lambda\hat{\text{k}})=0$
$2+3\lambda=0$
$-2=3\lambda$
$\lambda=\frac{-2}{3}$

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Question 31 Mark

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

$0$
$1$
$\frac{-2}{3}$
$\frac{-3}{2}$

Answer

$\frac{-2}{3}$

Solution:
Since, two non zero vector $\vec{\text{a}}\ \&\ \vec{\text{b}}$ are i.e.,
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$
$\vec{\text{a}}.\vec{\text{b}}=0$
$(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}).(2\hat{\text{i}}+\lambda\hat{\text{k}})=0$
$2+3\lambda=0$
$-2=3\lambda$
$\lambda=\frac{-2}{3}$

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Question 41 Mark

The ratio in which 2x + 3y + 5z = 1 divides the line joining the points (1, 0, -3) and (1, -5, 7) is:

5 : 3
3 : 2
2 : 1
1 : 3

Answer

5 : 3

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Question 51 Mark

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=$

$2\overrightarrow{\text{OG}}$
$4\overrightarrow{\text{OG}}$
$5\overrightarrow{\text{OG}}$
$3\overrightarrow{\text{OG}}$

Answer

$4\overrightarrow{\text{OG}}$

Solution:
Let us consider the point O as origin.
G is the mid-point of AC.

$\therefore\ \overrightarrow{\text{OG}}=\frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}2$
$2\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}\ \dots(1)$
Also, G is the mid-point BD
$\therefore\ \overrightarrow{\text{OG}}=\frac{\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}}2$
$2\overrightarrow{\text{OG}}=\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}\ \dots(2)$
On adding (1) and (2) we get,
$2\overrightarrow{\text{OG}}+2\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}$
$4\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}$
$\therefore\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\overrightarrow{\text{OG}}$

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Question 61 Mark

Choose the correct answer from the given four options.
If $|\vec{{\text{a}}}|=10,|\vec{{\text{b}}}|=2$ and $\vec{{\text{a}}}\cdot\vec{{\text{b}}}=12,$ then value of $|\vec{{\text{a}}}\times\vec{\text{b}}|$ is:

5.
10.
14.
16.

Answer

16.

Solution:
Here, $|\vec{{\text{a}}}|=10,|\vec{{\text{b}}}|=2$ and $\vec{{\text{a}}}\cdot\vec{\text{b}}=12$ [given]
$\therefore\vec{{\text{a}}}\cdot\vec{\text{b}}=|\vec{{\text{a}}}||\vec{{\text{b}}}|\cos\theta$
$12=10\times2\cos\theta$
$\Rightarrow\cos\theta=\frac{12}{20}=\frac{3}{5}$
$\Rightarrow\sin\theta=\sqrt{1-\cos\theta}$
$=\sqrt{1-\frac{9}{25}}$
$\sin\theta=\pm\frac{4}{5}$
$\therefore|\vec{{\text{a}}}\times\vec{{\text{b}}}|=|\vec{{\text{a}}}||\vec{{\text{b}}}||\sin\theta|$
$=10\times2\times\frac{4}{5}$
$=16$

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Question 71 Mark

If AD, BE and CF are $\triangle\text{ABC},$ then $\vec{\text{AD}}+\vec{\text{BE}}+\vec{\text{CF}}$

$\vec{0}$
1
0
2

Answer

$\vec{0}$

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Question 81 Mark

If O and O' are circumcenter and orthocenter of $\triangle{\text{ABC}}$ , then $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}$ equals,

$2\overrightarrow{\text{OO}'}$
$\overrightarrow{\text{OO}'}$
$\overrightarrow{\text{O}'\text{O}}$
$2\overrightarrow{\text{O}'\text{O}}$

Answer

$\overrightarrow{\text{OO}'}$

Solution:
Given: O be the circumcentre an O' be the orthocenter of $\triangle{\text{ABC}}$. Let G be the centroid of the triangle.
We know that O, G and H are collinear and by geometry $\overrightarrow{\text{O}'\text{G}}=2\overrightarrow{\text{OG}}$. This yields, $\overrightarrow{\text{O}'\text{O}}=\overrightarrow{\text{O}'\text{G}}+\overrightarrow{\text{GO}}=2\overrightarrow{\text{GO}}+\overrightarrow{\text{GO}}=3\overrightarrow{\text{GO}}$
In other words $\overrightarrow{\text{OO}'}=3\overrightarrow{\text{GO}}$
Since, $\overrightarrow{\text{OG}}=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
$\therefore\overrightarrow{\text{OO}'}=3\times\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
$=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}$

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Question 91 Mark

Which of the following represents collinear but not equal vectors:

a, c
b, d
b, m
Both (a) and (b)

Answer

a, c

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Question 101 Mark

Which of the following represents equal vectors:

a, c
b, d
b, c
m, d

Answer

b, d

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Question 111 Mark

Which of the following represents coinitial vector:

c, d
m, b
b, d
Both (a) and (b)

Answer

Both (a) and (b)

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Question 121 Mark

The magnitude of the vector 6i + 2j + 3k is equal to:

5
1
7
12

Answer

7

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Question 131 Mark

If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}} $ are three non-coplanar mutually perpendicular unit vectors, then $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big],$ is:

$\pm 1$
$0$
$-2$
$2$

Answer

$\pm1$

Solution:
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|\big|\vec{\text{c}}\big|\cos0^\circ$ or $\big|\vec{\text{a}}\times{\vec{\text{b}}}\big|\big|\vec{\text{c}}\big|\cos180^\circ$ $\big(\therefore\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other$)$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ or $-\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ $\big(\therefore\big|\vec{\text{c}}\big|=1,\cos0^\circ=1\text{ and }\cos180^\circ=-1\big)$ $$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin90^\circ$ or $-\big|\text{a}\big|\big|\vec{\text{b}}\big|\sin90$ $\big(\therefore\vec{\text{a}} \text{ is perpendicular to }\vec{\text{b}})$
$=1 \text{ or }-1$ $\big(\therefore\big|\vec{\text{a}}\big|=1 \text{ and }\big|\vec{\text{b}}\big|=1\big)$ $$
$=\pm1$

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Question 141 Mark

If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two unit vectors inclined at an angle $\theta$, such that $\big|\vec{\text{a}}+\vec{\text{b}}\big|<1,$ then:

$\theta<\frac{\pi}{3}$
$\theta>\frac{2\pi}{3}$
$\frac{\pi}{3}<\theta<\frac{2\pi}{3}$
$\frac{2\pi}{3}<\theta<\pi$

Answer

$\frac{2\pi}{3}<\theta<\pi$

Solution:
We have
$\big|\vec{\text{a}}+\vec{\text{b}}\big|<1$
$\Rightarrow\sqrt{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\theta<1}$
$\Rightarrow\sqrt{1^2+1^2+2\times1\times1\times\cos\theta<1}$
$\Rightarrow\sqrt{2+2\cos\theta}<1$
$\Rightarrow\sqrt{2(1+\cos\theta)}<1$
$\Rightarrow\sqrt{2\times2\cos^2\frac{\theta}{2}}<1$
$\Rightarrow2\big|\cos\frac{\theta}{2}\big|<1$
$\Rightarrow\big|\cos\frac{\theta}{2}\big|<\frac{1}{2}$
$\Rightarrow\frac{\pi}{3}<\frac{\theta}{2}<\frac{2\pi}{3}$
$\Rightarrow\frac{2\pi}{3}<\theta<\frac{4\pi}{3}$
But here $\theta$ cannot be more than $\pi.$

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Question 151 Mark

If $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}},$ then $\vec{\text{a}}\times\vec{\text{b}}$ is:

$10\hat{\text{i}}+2\hat{\text{j}}+11\hat{\text{k}}$
$10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
$10\hat{\text{i}}-3\hat{\text{j}}+11\hat{\text{k}}$
$10\hat{\text{i}}-2\hat{\text{j}}-10\hat{\text{k}}$

Answer

$10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$

Solution:
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&-1\\1&4&-2 \end{vmatrix}$
$=10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$

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Question 161 Mark

Choose the correct answer from the given four options.
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are unit vectors such that $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0,$ then the value of $\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}$ is:

1.
3.
$-\frac{3}{2}.$
None of these.

Answer

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

Solution:
We have $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})=0$
$\Rightarrow\vec{\text{a}}^2+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}^2+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{b}}+\vec{\text{c}}^2=0$
$\Rightarrow\vec{\text{a}}^2+\vec{\text{b}}^2+\vec{\text{c}}^2+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$ $[\because\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}},\vec{\text{b}}\cdot\vec{\text{c}}=\vec{\text{c}}\cdot\vec{\text{b}}\text{ and }\vec{\text{c}}\cdot\vec{\text{a}}=\vec{\text{a}}\cdot\vec{\text{c}}]$
$\Rightarrow1+1+1+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$
$\Rightarrow\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}=-\frac{3}{2}$

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Question 171 Mark

Two or more vectors having the same initial point are:

Coinitial vectors
Colinear vectors
Equal vectors
Cannot say

Answer

Coinitial vectors

Solution:
Two or more vectors having same initial points are known as co-initial vectors.

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MCQ 181 Mark

If the curve $ay + x^2 = 7$ and $x^3 = y,$ cut orthogonally at $(1, 1)$ then the value of $a$ is$:$

A
$1$
B
$0$
C
$-6$
✓
$6$

Answer

Correct option: D.

$6$

$6$

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Question 191 Mark

Let G be the centroid of $\triangle{\text{ABC}}$. if $\overrightarrow{\text{AB}}=\vec{\text{a}},\overrightarrow{\text{AC}}=\vec{\text{b}}$, then the bisector $\overrightarrow{\text{AG}}$, in terms of $\vec{\text{a}}$ and $\vec{\text{b}}$ is,

$\frac{2}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$\frac{1}6\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$\frac{1}2\big(\vec{\text{a}}+\vec{\text{b}}\big)$

Answer

$\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$

Solution:
Taking A as origin. Then, position vector of A, B and C are $\vec0,\vec{\text{a}}$ and $\vec{\text{b}}$ respectively. Then,
Centroid G has position vector $\frac{\vec0+\vec{\text{a}}+\vec{\text{b}}}3=\frac{\vec{\text{a}}+\vec{\text{b}}}3$
Therefore, $\text{AG}=\frac{\vec{\text{a}}+\vec{\text{b}}}3-\vec0=\frac{\vec{\text{a}}+\vec{\text{b}}}3$

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Question 201 Mark

If $\theta$ is an acute angle and the vector $(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}$ is perpendicular to the vector $\hat{\text{i}}-\sqrt{3}\hat{\text{j}},$

$\frac{\pi}{6}$
$\frac{\pi}{5}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$

Answer

$\frac{\pi}{3}$

Solution:
The given vectors are perpendicular. so, their dot product is zero.
$\big[(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}\big].\big(\hat{\text{j}}-\sqrt{3}\hat{\text{j}}\big)=0$
$\Rightarrow\sin\theta-\sqrt{3}\cos\theta=0$
$\Rightarrow\sin\theta=\sqrt{3}\cos\theta$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\theta=\frac{\pi}{3}$ (because $\theta$ is acute)

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Question 211 Mark

Choose the correct answer from the given four options.
If $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2,$ then the range of $|\lambda\vec{\text{a}}|$ is:

[0,8]
[-12,8]
[0,12]
[8,12]

Answer

[0,12]

Solution:
We have $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2$
Now $|\lambda\vec{\text{a}}|=|\lambda||\vec{\text{a}}|=4|\lambda|$
Now $-3\leq\lambda\leq2$
$\Rightarrow0\leq|\lambda|\leq3$
$\Rightarrow0\leq4|\lambda|\leq12$
$\Rightarrow0\leq|\lambda\vec{\text{a}}|\leq12$

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Question 221 Mark

For any three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ the expression $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$ equals:

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}]}^2$
None of these

Answer

None of these

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

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Question 231 Mark

The system of vectors i, j, k is:

Orthogonal
Collinear
Coplanar
None of these

Answer

Orthogonal

Solution:
Since i, j, k represent unit vector in the direction of X,Y and Z axis respectively.
Therefore, they are orthogonal.

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Question 241 Mark

If ABCDEF is a regular hexagon, then $\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}+\overrightarrow{\text{FC}}$ equals,

$2\overrightarrow{\text{AB}}$
$\vec0$
$3\overrightarrow{\text{AB}}$
$4\overrightarrow{\text{AB}}$

Answer

$4\overrightarrow{\text{AB}}$

Solution:

$\overrightarrow{\text{AD}}=2\overrightarrow{\text{BC}}$
$\overrightarrow{\text{EB}}=2\overrightarrow{\text{FA}}$
$\overrightarrow{\text{FC}}=2\overrightarrow{\text{AB}}$
$\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=2\Big(\overrightarrow{\text{BC}}+\overrightarrow{\text{FA}}\Big)$
$=2\Big(\overrightarrow{\text{AO}}+\overrightarrow{\text{FA}}\Big)$ $\Big(\because\ \overrightarrow{\text{BC}}=\overrightarrow{\text{AO}}\Big)$
In triangle AOF,
$\overrightarrow{\text{FA}}+\overrightarrow{\text{AO}}+\overrightarrow{\text{FO}}=0$
$\therefore\ \overrightarrow{\text{FA}}+\overrightarrow{\text{AO}}=-\overrightarrow{\text{FO}}$
$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=-2\overrightarrow{\text{FO}}$
And $\overrightarrow{\text{AB}}=-\overrightarrow{\text{FO}}$
$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=2\overrightarrow{\text{AB}}$
$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}+\overrightarrow{\text{FC}}=2\overrightarrow{\text{AB}}+2\overrightarrow{\text{AB}}$
$=4\overrightarrow{\text{AB}}$

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Question 251 Mark

A vector whose initial and terminal points coincide, is:

Zero Vector
Equal Vectors
Null Vector
Unit Vector

Answer

Zero Vector

Solution:
The vector whose initial and terminals points are coincide has the length 0.
we call it to be a zero vector and the zero vector no has the particular direction,
so that it can be assigned in any direction.

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Question 261 Mark

Forces $3\overrightarrow{\text{OA}},\ 5\overrightarrow{\text{OB}}$ act along OA and OB. If their resultant passes through C on AB, then,

C is a mid-point of AB.
C divides AB in the ratio 2 : 1
3AC = 5CB
2AC = 3CB

Answer

3AC = 5CB

Solution:
Draw ON, the perpendicular to the line AB

Let $\vec{\text{i}}$ be the unit vector along ON
The resultant force $\vec{\text{R}}=3\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}\ \dots(1)$
The angles between $\vec{\text{i}}$ and the forces $\vec{\text{R}},\ 3\overrightarrow{\text{OA}},\ 5\overrightarrow{\text{OB}}$ are $\angle\text{CON},\ \angle\text{AON},\ \angle\text{BON}$ respectively.
$\vec{\text{R}}.\vec{\text{i}}=3\overrightarrow{\text{OA}}.\vec{\text{i}}+5\overrightarrow{\text{OB}}.\vec{\text{i}}$
$\Rightarrow\text{R}.1.\cos\angle\text{CON}\\=3\overrightarrow{\text{OA}}.1.\cos\angle\text{AON}+5\overrightarrow{\text{OB}}.1.\cos\angle{\text{BON}}$
$\text{R}.\frac{\text{ON}}{\text{OC}}=3\text{OA}\times\frac{\text{ON}}{\text{OA}}+5\text{OB}\frac{\text{ON}}{\text{OB}}$
$\frac{\text{R}}{\text{OC}}=(3+5)$
$\text{R}=8\overrightarrow{\text{OC}}$
We know that,
$\overrightarrow{\text{OA}}=\overrightarrow{\text{OC}}+\overrightarrow{\text{CA}}$
$\Rightarrow3\overrightarrow{\text{OA}}=3\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}\ \dots(\text{i})$
$\overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}+\overrightarrow{\text{CB}}$
$\Rightarrow5\overrightarrow{\text{OB}}=5\overrightarrow{\text{OC}}+5\overrightarrow{\text{CB}}\ \dots(\text{ii})$
On adding (i) and (ii) we get,
$3\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$
$\vec{\text{R}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$
$8\overrightarrow{\text{OC}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$
$\Big|3\overrightarrow{\text{AC}}\Big|=\Big|5\overrightarrow{\text{CB}}\Big|$
$\Rightarrow3\overrightarrow{\text{AC}}=5\overrightarrow{\text{CB}}$

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Question 271 Mark

vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are inclined at angle $\theta=120^\circ.$ if $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=2,$ then $\big[\big(\vec{\text{a}}+3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)\big]^2$ is equal to:

300
325
275
225

Answer

300

Solution:
$\big(\vec{\text{a}}+3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)$
$=3\big(\vec{\text{a}}\times\vec{\text{a}}\big)-\vec{\text{a}}\times\vec{\text{b}}+9\big(\vec{\text{b}}\times\vec{\text{a}}\big)-3\big(\vec{\text{b}}\times\vec{\text{b}}\big)$
$=3(0)-\vec{\text{a}}\times\vec{\text{b}}-9\big(\vec{\text{a}}\times\vec{\text{b}}\big)-3(0)$
$=-10\big(\vec{\text{a}}\times\vec{\text{b}}\big)$
Now,
$\big|\big(\vec{\text{a}}\times3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)\big|^2$
$=\big|-10\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2$
$=100\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2$
$=100|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2120$
$=100(1)^2(2)^2\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=400\times\frac{3}{4}$
$=300$

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Question 281 Mark

The Polygon Law of Vector Addition is simply an extension of:

Parallelogram Law of Vector Addition
Triangular Law of Vector Addition
Both A and B
None of the above

Answer

Triangular Law of Vector Addition

Solution:
The Polygon Law of Vector Addition is simply an extension of Triangular Law of Vector Addition.
Here, we take into consideration more than two sides unlike in triangular law of vector addition.

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Question 291 Mark

The Polygon Law of Vector Addition is simply an extension of ____________?

Parallelogram Law of Vector Addition
Triangular Law of Vector Addition
Both A and B
None of the above

Answer

Triangular Law of Vector Addition

Solution:
The Polygon Law of Vector Addition is simply an extension of Triangular Law of Vector Addition.
Here we take into consideration more than two sides unlike in triangular law of vector addition.

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Question 301 Mark

If two or more vectors are parallel to the same line, irrespective of their magnitudes and directions, then they are:

Co-initial vectors
Collinear vectors
Equal vectors
Unit vectors

Answer

Collinear vectors

Solution:
Collinear vectors are two or more vectors which are parallel to the same line irrespective of their magnitude and direction.

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Question 311 Mark

If $\vec{\text{a}}$ and $\vec{\text{b}}$ be two unit vectors and $\theta$ the angle between them, than $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if $\theta=$

$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{2\pi}{3}$

Answer

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

Solution:
We have
$|\vec{\text{a}}|=1$ and $\big|\vec{\text{b}\big|}=1$
Now, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}\big|}^2+2\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow1+1+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=1$
$\Rightarrow2+2\cos\theta=1$
$\Rightarrow2\cos\theta=-1$
$\Rightarrow\cos\theta=\frac{-1}{2}$
$\Rightarrow\theta=\frac{2\pi}{3}$

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Question 321 Mark

The vector component of $\vec{\text{b}}$ perpendicular to $\vec{\text{a}}$ is:

$\big(\vec{\text{b}}.\vec{\text{c}}\big)\vec{\text{a}}$
$\frac{\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)}{|\vec{\text{a}}|^2}$
$\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)$
None of these

Answer

$\frac{\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)}{|\vec{\text{a}}|^2}$

Solution:
The vector component of $\vec{\text{b}}$ perpendicular to $\vec{\text{a}}$ is
$\frac{\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)}{|\vec{\text{a}}|^2}$

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MCQ 331 Mark

Which of the below given is a vector quantity$:$

A
$8\ kg$
B
$4$ seconds
✓
$6$ Newton
D
$90\ cm^3$

Answer

Correct option: C.

$6$ Newton

$6$ Newton is a vector quantity as it is a force. Force is a vector quantity which has both magnitude and direction.

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Question 341 Mark

What is the value of x and y, if 2i + 3j = xi + yj:

4, 9
3, 2
2, 3
0, 0

Answer

2, 3

Solution:
2i + 3j = xi + yj
On comparing the two equations, we have,
x = 2 and y = 3

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Question 351 Mark

If the points A and B are (1, 2, -1), and (2, 1, -1) respectively, then $ \vec{ \text{AB }} $ is:

$\hat{\text{i}}+\hat{\text{j}}$
$\hat{\text{i}}-\hat{\text{j}}$
$2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$

Answer

$\hat{\text{i}}-\hat{\text{j}}$

Solution:
$ \vec{ \text{AB}}=\langle{2-1, 1-2,-1+1}\rangle=\langle{1, -1, 0}\rangle\therefore \vec{ \text{AB}}=\hat{\text{i}}-\hat{\text{j}}$

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Question 361 Mark

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=$

$\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2$
$\big|\vec{\text{a}}\big|^2+\big|\vec{\text{b}}\big|^2$
$2\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

Answer

$\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

Solution:
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$
$=\big|\vec{\text{a}}\big|^2\big|{\text{b}}\big|^2\sin^2\theta+\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

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Question 371 Mark

Choose the correct answer from the given four options.
Find the value of $\lambda$ such that the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ are orthogonal:

$0$
$1$
$\frac{3}{2}$
$-\frac{5}{2}$

Answer

$-\frac{5}{2}$

Solution:
Given two non-zero vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are orthogonal
$\therefore\ \vec{\text{a}}\cdot\vec{\text{b}}=0$
$\therefore\ (2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}+\hat{3\text{k}})=0$
$\Rightarrow2+2\lambda+3=0$
$\Rightarrow\lambda=-\frac{5}{2}$

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Question 381 Mark

The value of $\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big],$ where $\big|\vec{\text{a}}\big|=1,\big|\vec{\text{b}}\big|=5,\big|\vec{\text{c}}\big|=3,$ is:

0
1
6
None of these.

Answer

0

Solution:
We have
$\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big]$
$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big).\big(\vec{\text{c}}.\vec{\text{a}}\big)$ (By definition of scalar triple product)
$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{b}}-\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{b}}\times{\vec{\text{b}}}-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}-0-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}-\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{a}}-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{c}}+\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{a}}+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{c}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)\\.\vec{\text{a}}\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]-\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{c}}\big]+\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{a}}\big]+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{c}}\big]-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0+0+0-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$
$=0$

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Question 391 Mark

Choose the correct answer from the given four options.
For any vector $\vec{\text{a}},$ the value of $(\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ is:

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

Answer

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

Solution:
Let $\vec{\text{a}}^2=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$\therefore\vec{\text{a}}^2=\text{x}^2+\text{y}^2+\text{z}^2$
$\therefore\ \vec{\text{a}}\times\vec{\text{i}}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$
$=\hat{\text{i}}[0]-\hat{\text{j}}[-\text{z}]+\hat{\text{k}}[-\text{y}]$
$=\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$
$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})=(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})$
$=\text{y}^2+\text{z}^2$
Similarly, $(\vec{\text{a}}\times\hat{\text{j}})^2=\text{x}^2+\text{z}^2$
And $(\vec{\text{a}}\times\hat{\text{k}})^2=\text{x}^2+\text{y}^2$
$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ $=\text{y}^2+\text{z}^2+\text{x}^2+\text{z}^2+\text{x}^2+\text{y}^2$
$2(\text{x}^2+\text{y}^2+\text{z}^2)=2\vec{\text{a}}^2$

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Question 401 Mark

Which of the following is not a vector quantity:

Speed
Density
Force
Velocity

Answer

Density

Solution:
Density is a scalar quantity as it has only magnitude but no direction. Speed, force, velocity has both magnitude and direction.
$\therefore$ They all are vectors.

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Question 411 Mark

If one point on the vector 2i − 4j − k is (2, 1, 3), the other point is?

(−4, 3, 2)
(4, −3, −2)
(3, 2, 1)
(4, −3, 2)

Answer

(4, −3, 2)

Solution:
Let the other point on the vector 2i − 4j − k be (x, y, z).
Given point on the vector is (2, 1, 3).
Thus the vector is represented as
(2, − 4, − 1) = (x, y, z) − (2, 1, 3)
(2, − 4, − 1) = (x − 2, y − 1, z − 3)
Equating the corresponding points we get 2 = x − 2, −4 = y − 1,−1 = z − 3
⇒ x = 4, y = −3, z = 2

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Question 421 Mark

What is the magnitude of vector -3i + 5j:

$\sqrt{34}$
$\sqrt{32}$
$\sqrt{8}$
$\sqrt{16}$

Answer

$\sqrt{34}$

Solution:
Vector, V = -3i + 5j
Magnitude of the vector, V
$\mid\text{v}\mid=\sqrt{{(-3)^2}+5^2}=\sqrt{(9+25)}=\sqrt{34}$

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Question 431 Mark

If $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0},|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5,|\vec{\text{c}}|=7,$then the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is:

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

Answer

$\frac{\pi}{3}$

Solution:
Given, $|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5$ and $|\vec{\text{c}}|=7\dots(1)$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given that
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=-\vec{\text{c}}$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=|-\vec{\text{c}}|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2-|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=7^2-3^3-5^2$ [using (1)]
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=15$
$\Rightarrow2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=15$
$\Rightarrow2(3)(5)\cos\theta=15$ [using (1)]
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}$

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Question 441 Mark

If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ then a unit vector normal to the vectors $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}$ is:

$\hat{\text{i}}$
$\hat{\text{j}}$
$\hat{\text{k}}$
$\text{None of these}$

Answer

$\hat{\text{i}}$

Solution:
$\vec{\text{a}}+\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}-\vec{\text{c}}=0\hat{\text{i}}-0\hat{\text{j}}+3\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&3&1\\0&0&3 \end{vmatrix}$
$=9\hat{\text{i}}$
$\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=9|\hat{\text{i}}|$
$=9(1)$
$=9$
Unit vector perpendicular to both $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)}{\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|}$
$=\frac{9\hat{\text{i}}}{9}$
$=\hat{\text{i}}$

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Question 451 Mark

A zero vector has:

Any direction
Many directions
No direction
None of these

Answer

Any direction

Solution:
A zero or null vector have any or no direction.

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Question 461 Mark

Let a, b and c be vectors with magnitudes 3, 4 and 5 respectively and a + b + c = 0, then the values of a.b + b.c + c.a is:

47
25
50
-25

Answer

-25

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Question 471 Mark

The values of x for which the angle between $\vec{\text{a}}=2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}$is obtuse and the angle between $\vec{\text{b}}$ and the z-axis is acute and less than $\frac{\pi}{6}$ are:

$\text{x}>\frac{1}{2}$ or $\text{x}<0$
$0<\text{x}<\frac{1}{2}$
$\frac{1}{2}<\text{x}<15$
$\phi$

Answer

$0<\text{x}<\frac{1}{2}$

Solution:
$\vec{\text{a}}=2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}, \vec{\text{b}}=7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}$
Let the angle between vector a and vector b be A.
$\therefore\cos\text{A}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{\big(2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}\big).\big(7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big)}{\big|2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}\big|\big|7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big|}$
$=\frac{14\text{x}^2-8\text{x}+\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{49+4+\text{x}^2}}$
$=\frac{14\text{x}^2-8\text{x}+\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{53+\text{x}^2}}$
Now, $\angle\text{A}$ is an obtuse angle.
$\therefore\cos\text{A}<0$
$\Rightarrow\frac{14\text{x}^2-7\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{53+\text{x}^2}}<0$
$\Rightarrow14\text{x}^2-7\text{x}<0$
$\Rightarrow2\text{x}^2-\text{x}<0$
$\Rightarrow\text{x}(2\text{x}-1)<0$
$\Rightarrow\text{x}<0\ \&\ 2\text{x}-1>0$ or $\text{x}>0\ \&\ 2\text{x}-1<0$
$\Rightarrow\text{x}<0\ \&\ \text{x}>\frac{1}{2}$ or $\text{x}>0\ \&\ \text{x}<\frac{1}{2}$
$\Rightarrow\text{x}>0\ \&\ \text{x}<\frac{1}{2}$ (As there cannot be any number less than zero and greater than $\frac{1}{2}$)
$\Rightarrow\text{x}\in(0,\frac{1}{2})\dots(1)$
Let the equation of the z-axis be $\text{z}\hat{\text{k}}.$
And let the angle between $\vec{\text{b}}$ and z-axis be B.
$\therefore\cos\text{B}=\frac{\big(7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big).\big(\text{z}\hat{\text{k}}\big)}{\big|7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big|\big|\text{z}\hat{\text{k}}\big|}$
$=\frac{\text{xz}}{\text{z}\sqrt{49+4+\text{x}^2}}$
$=\frac{\text{x}}{\sqrt{53+\text{x}^2}}$
Now, angle B is acute and less than $\frac{\pi}{6}.$
$\therefore0<\frac{\text{x}}{\sqrt{53+\text{x}^2}}<\cos\frac{\pi}{6}$
$\Rightarrow0<\text{x}<\frac{\sqrt{3}}{2}\sqrt{53+\text{x}^2}\dots(2)$
From (1) and (2) we get
$0<\text{x}<\frac{1}{2}$

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Question 481 Mark

Let $\vec{\text{a}}$ and $\vec{\text{b}}$ be two unit vectors and a be the angle between them. Then, $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if:

$\text{a}=\frac{\pi}{4}$
$\text{a}=\frac{\pi}{3}$
$\text{a}=\frac{2\pi}{3}$
$\text{a}=\frac{\pi}{2}$

Answer

$\text{a}=\frac{2\pi}{3}$

Solution:
$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors.
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$
Now,
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\text{a}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=\cos\text{a}\dots(2)$
[using (1)]
Given that
$\Big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow1+1+2\cos\text{a}=1$ [using (1) and (2)]
$\Rightarrow2+2\cos\text{a}=1$
$\Rightarrow2\cos\text{a}=-1$
$\Rightarrow\cos\text{a}=\frac{-1}{2}$
$\Rightarrow\text{a}=\frac{2\pi}{3}$

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Question 491 Mark

Choose the correct answer from the given four options.
The position vector of the point which divides the join of points $2\vec{\text{a}}-3\vec{\text{b}}$ and $\vec{\text{a}}+\vec{\text{b}}$ in the ratio 3 : 1 is:

$\frac{3\vec{\text{a}}-2\vec{\text{b}}}{2}$
$\frac{7\vec{\text{a}}-8\vec{\text{b}}}{4}$
$\frac{3\vec{\text{a}}}{4}$
$\frac{5\vec{\text{a}}}{4}$

Answer

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

Solution:
Let the given points be $\text{A}(2\vec{\text{a}}-3\vec{\text{b}})$ and $\text{B}(\vec{\text{a}}+\vec{\text{b}})$
Let C divides AB in ratio 3 : 1
Now the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally, is given by $\frac{\text{m}\vec{\text{q}}+\text{n}\vec{\text{p}}}{\text{m}+\text{n}}$
$\therefore$ Position vector $\text{C}=\frac{3(\vec{\text{a}}+\vec{\text{b}})+1(\vec{2\text{a}}-3\vec{\text{b}})}{3+1}$
$\Rightarrow\text{C}=\frac{5\vec{\text{a}}}{4}$

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Question 501 Mark

The vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}$ is to be written as the sum of a vector $\vec{\alpha}$ parallel to $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$ and a vector $\vec{\beta}$ perpendicular to $\vec{\text{a}}.$ Then $\vec{\alpha}=$

$\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
$\frac{2}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
$\frac{1}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
$\frac{1}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$

Answer

$\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$

Solution:
Let:
$\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\beta}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
Now,
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}=\vec{\alpha}+\vec{\beta}$ (Given)
$\Rightarrow3\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}=(\text{a}_1+\text{b}_1)\hat{\text{i}}+(\text{a}_2+\text{b}_2)\hat{\text{j}}+(\text{a}_3+\text{b}_3)\hat{\text{k}}$
$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2+\text{b}_2=0;\text{a}_3+\text{b}_3=4$
$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2=-\text{b}_2;\text{a}_3+\text{b}_3=4\dots(1)$
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$ (Given)
Also, $\vec{\alpha}$ is parallrl to $\vec{\text{a}}.$
$\Rightarrow\vec{\alpha}\times\vec{\text{a}}=\vec{0}$
$\Rightarrow\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\1&1&0\end{vmatrix}=\vec{0}$
$\Rightarrow-\text{a}_3\hat{\text{i}}+\text{a}_3\hat{\text{j}}+(\text{a}_1-\text{a}_2)\hat{\text{k}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow\text{a}_3=0;\text{a}_1-\text{a}_2=0$
$\Rightarrow\text{a}_3=0;\text{a}_1=\text{a}_2\dots(2)$
Since $\vec{\beta}$ is perpendicular to $\vec{\text{a}},$ we get
$\Rightarrow\vec{\beta}.\vec{\text{a}}=0$
$\Rightarrow\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big).\big(\hat{\text{i}}.\hat{\text{j}}\big)=0$
$\Rightarrow\text{b}_1+\text{b}_2=0$
$\Rightarrow\text{b}_1=-\text{b}_2\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}_1=\frac{3}{2};\text{a}_2=\frac{3}{2};\text{a}_3=0$
$\therefore\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$=\frac{3}{2}\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+0\hat{\text{k}}$
$=\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$

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