MCQ 2011 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
$-3$
AnswerCorrect option: C. $\frac{-3}{2}$
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$)$
$\Rightarrow |\vec{a}+\vec{b}+\vec{c}|^2=0$
$\Rightarrow |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow 3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow (\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}$
View full question & answer→MCQ 2021 Mark
If $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+p \hat{j}+q \hat{k})=\overrightarrow{0}$, then which of the following is true?
- A
$p=6, q=27$
- ✓
$p=3, q=\frac{27}{2}$
- C
$p=6, q=\frac{27}{2}$
- D
$p=3, q=27$
AnswerCorrect option: B. $p=3, q=\frac{27}{2}$
(b) : Given,
$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & p & q\end{array}\right|=0$
$
\begin{array}{ll}
\Rightarrow & \hat{i}(6 q-27 p)-\hat{j}(2 q-27)+\hat{k}(2 p-6)=0 \\
\Rightarrow & 6 q-27 p=0,2 q-27=0 \text { and } 2 p-6=0 \\
\Rightarrow & q=\frac{27}{2} \text { and } p=3 .
\end{array}
$
View full question & answer→MCQ 2031 Mark
If $|\vec{a}|=4$ and $-3 \leq \lambda \leq 3$, then which of the following is the range of $|\lambda \vec{a}|$ ?
(i) $[0,8]$
(ii) $[-12,8]$
(iii) $[0,12]$
Answer(b) : We have, $-3 \leq \lambda \leq 3 \Rightarrow|\lambda| \leq 3$
Now, $|\lambda||\vec{a}| \leq 3|\vec{a}| \Rightarrow|\lambda \vec{a}| \leq 12$
$\therefore \quad$ Range of $|\lambda \vec{a}|$ is $[0,12]$
View full question & answer→MCQ 2041 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
AnswerCorrect option: D. $\frac{5 \vec{a}}{4}$
Given points are $2 \vec{a}-3 \vec{b}$ and $\vec{a}+\vec{b}$.
Given ratio $=3: 1$
$\therefore$ Required vector $=\frac{(2 \vec{a}-3 \vec{b}) \times 1+(\vec{a}+\vec{b}) \times 3}{3+1}$
$=\frac{2 \vec{a}-3 \vec{b}+3 \vec{a}+3 \vec{b}}{4}$
$=\frac{5}{4} \vec{a}$
View full question & answer→MCQ 2051 Mark
The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4 , respectively and $\vec{a} \cdot \vec{b}=2 \sqrt{3}$ is
- A
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{3}$
- C
$\frac{\pi}{2}$
- D
$\frac{5 \pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{3}$
(b) : We have $\vec{a} \cdot \vec{b}=2 \sqrt{3},|\vec{a}|=\sqrt{3},|\vec{b}|=4$
Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.
$
\begin{array}{ll}
\therefore & \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\
\Rightarrow & 2 \sqrt{3}=\sqrt{3} \cdot 4 \cdot \cos \theta \quad \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
$
View full question & answer→MCQ 2061 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
$7,5,-4$
AnswerCorrect option: B. $7,-5,4$
We have, $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}, \vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$
$\therefore 3 \vec{a}+2 \vec{b}=3(\hat{i}+\hat{j}-2 \hat{k})+2(2 \hat{i}-4 \hat{j}+5 \hat{k})$
$=(3 \hat{i}+3 \hat{j}-6 \hat{k})+(4 \hat{i}-8 \hat{j}+10 \hat{k})$
$=7 \hat{i}-5 \hat{j}+4 \hat{k}$
$\therefore$ The direction ratios of the vector $3 \vec{a}+2 \vec{b}$ are $7,-5,4$.
View full question & answer→MCQ 2071 Mark
Area of a parallelogram whose adjacent sides are represented by the vectors $2 \hat{i}-3 \hat{k}$ and $4 \hat{j}+2 \hat{k}$ is
- ✓
$4 \sqrt{14}$ sq. units
- B
$2 \sqrt{7}$ sq. units
- C
$4 \sqrt{7}$ sq. units
- D
$4 \sqrt{19}$ sq. units
AnswerCorrect option: A. $4 \sqrt{14}$ sq. units
Let $\vec{a}=2 \hat{i}-3 \hat{k}$ and $\vec{b}=4 \hat{j}+2 \hat{k}$
The area of a parallelogram with $\vec{a}$ and $\vec{b}$ as its adjacent sides is given by $|\vec{a} \times \vec{b}|$.
Now, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -3 \\ 0 & 4 & 2\end{array}\right|=12 \hat{i}-4 \hat{j}+8 \hat{k}$
$ \therefore |\vec{a} \times \vec{b}|=\sqrt{(12)^2+(-4)^2+(8)^2}=\sqrt{144+16+64}$
$\quad=\sqrt{224}=4 \sqrt{14} \text { sq. units. } $
View full question & answer→MCQ 2081 Mark
The value of $p$ for which $p(\hat{i}+\hat{j}+\hat{k})$ is a unit vector is
- A
- ✓
$\frac{1}{\sqrt{3}}$
- C
- D
$\sqrt{3}$
AnswerCorrect option: B. $\frac{1}{\sqrt{3}}$
(b) : Let $\vec{a}=(\hat{i}+\hat{j}+\hat{k})$
So, unit vector of $\vec{a}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{1+1+1}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
$\therefore \quad$ The value of $p$ is $\frac{1}{\sqrt{3}}$.
View full question & answer→MCQ 2091 Mark
If $|\vec{a}|=10,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$, then the value of $|\vec{a} \times \vec{b}|$ is
Answer$|\vec{a}|=10,|\vec{b}|=2, \vec{a} \cdot \vec{b}=12$
We know that, $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
$\Rightarrow 12=10 \times 2 \cos \theta$
$\Rightarrow \cos \theta=\frac{3}{5}$
$\therefore \sin \theta=\frac{4}{5}$
Now, $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=10 \times 2 \times \frac{4}{5}=16$
View full question & answer→MCQ 2101 Mark
Let $\vec{a}$ and $\vec{b}$ are unit vectors enclosing an angle $\theta$ and $|\vec{a}+\vec{b}|<1$. Which of the following is true?
$(i) \theta=\frac{\pi}{2}$
$(ii) \theta<\frac{\pi}{3}$
$(iii)\pi \geq \theta>\frac{2 \pi}{3}$
$(iv) \cos \theta<-\frac{1}{2}$
- A
Only $(i)$
- B
Only $(ii)$
- ✓
Only $(iii)$ and $(iv)$
- D
AnswerCorrect option: C. Only $(iii)$ and $(iv)$
$|\vec{a}+\vec{b}|<1$
$\Rightarrow|\vec{a}+\vec{b}|^2<1$
$\Rightarrow|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}<1$
$\Rightarrow 1+1+2 \vec{a} \cdot \vec{b}<1 [\because|\vec{a}|=|\vec{b}|=1]$
$\Rightarrow \vec{a} \cdot \vec{b}<-\frac{1}{2}$
$\Rightarrow|\vec{a}||\vec{b}| \cos \theta<-\frac{1}{2}$
$\Rightarrow 1 \times 1 \times \cos \theta<-\frac{1}{2}$
$\Rightarrow \cos \theta<-\frac{1}{2}$
$\Rightarrow-1 \leq \cos \theta<-\frac{1}{2}$
$\Rightarrow \pi \geq \theta>\frac{2 \pi}{3}$
View full question & answer→MCQ 2111 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 $(\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} \cdot(\vec{b} \times \vec{c})$.
AnswerCorrect 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}
$
View full question & answer→MCQ 2121 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
AnswerCorrect option: B. $\sqrt{2}$
Clearly, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.
$\Rightarrow \vec{a} \cdot \vec{b}=0$
$\therefore|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1+1+0=2$
$\Rightarrow|\vec{a}+\vec{b}|=\sqrt{2}$
View full question & answer→MCQ 2131 Mark
$(\hat{i}+\hat{j}) \times(\hat{j}+\hat{k}) \cdot(\hat{k}+\hat{i})$ is equal to
Answer$(\hat{i}+\hat{j}) \times(\hat{j}+\hat{k}) \cdot(\hat{k}+\hat{i})=(\hat{i} \times \hat{j}+\hat{i} \times \hat{k}+\hat{j} \times \hat{k}) \cdot(\hat{k}+\hat{i})$
$=(\hat{k}-\hat{j}+\hat{i}) \cdot(\hat{k}+\hat{i})=\hat{k} \cdot \hat{k}+\hat{i} \cdot \hat{i} \quad(\because \hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0)$
$=|\hat{k}|^2+|\hat{i}|^2=1+1=2$
View full question & answer→MCQ 2141 Mark
If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$, then $|\vec{a} \times \vec{b}|$ is equal to
- ✓
$\sqrt{507}$
- B
$\sqrt{506}$
- C
$\sqrt{508}$
- D
$\sqrt{509}$
AnswerCorrect option: A. $\sqrt{507}$
(a): We have, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2\end{array}\right|$
$
=\hat{i}(-2-15)-(-4-9) \hat{j}+(10-3) \hat{k}=-17 \hat{i}+13 \hat{j}+7 \hat{k}
$
Hence, $|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+(7)^2}=\sqrt{507}$
View full question & answer→MCQ 2151 Mark
Let $\vec{a}$ and $\vec{b}$ are non-collinear. If $\vec{c}=(x-2) \vec{a}+\vec{b}$ and $\vec{d}=(2 x+1) \vec{a}-\vec{b}$ are collinear, then find the value of $x$.
- A
$\frac{2}{3}$
- B
$\frac{-1}{3}$
- C
$\frac{-2}{3}$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
We have, $\vec{c}=(x-2) \vec{a}+\vec{b}, \vec{d}=(2 x+1) \vec{a}-\vec{b}$ are collinear, then $\vec{c}=m \vec{d}$ where $m$ is any scalar.
$\Rightarrow \quad(x-2) \vec{a}+\vec{b}=m((2 x+1) \vec{a}-\vec{b})$
$\Rightarrow \quad-m=1 $
$\Rightarrow m=-1$ and $m(2 x+1)=x-2 $
$\Rightarrow-2 x-1=x-2 $
$\Rightarrow x=\frac{1}{3}$
View full question & answer→MCQ 2161 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
AnswerCorrect 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 $\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)$
View full question & answer→MCQ 2171 Mark
If $\vec{u}=\hat{i}+2 \hat{j}, \vec{v}=-2 \hat{i}+\hat{j}$ and $\vec{w}=4 \hat{i}+3 \hat{j}$, then find scalars $x$ and $y$ such that $\vec{w}=x \vec{u}+y \vec{v}$.
- A
$x=4, y=-2$
- ✓
$x=2, y=-1$
- C
$x=3, y=5$
- D
$x=-5, y=2$
AnswerCorrect option: B. $x=2, y=-1$
We have, $\vec{w}=x \vec{u}+y \vec{v}$
$\Rightarrow 4 \hat{i}+3 \hat{j}=x(\hat{i}+2 \hat{j})+y(-2 \hat{i}+\hat{j})$
$\Rightarrow \quad(x-2 y-4) \hat{i}+(2 x+y-3) \hat{j}=\overrightarrow{0}$
$\Rightarrow x-2 y-4=0$ and $2 x+y-3=0$
$\Rightarrow x=2$ and $ y=-1$
View full question & answer→MCQ 2181 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
AnswerGiven, $|\vec{a}|=|\vec{b}|, \theta=60^{\circ}$ and $\vec{a} \cdot \vec{b}=\frac{9}{2}$
Now, $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\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 $
$\therefore|\vec{a}|=|\vec{b}|=3$
View full question & answer→MCQ 2191 Mark
If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}+2 \hat{k}$, then the value of $\vec{a} \cdot(\vec{b} \times \vec{c})$ is
AnswerHere, $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}+2 \hat{k}$
Now, $\vec{b} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 3 & 1 & 2\end{array}\right|=3 \hat{i}+5 \hat{j}-7 \hat{k}$
$\therefore \vec{a} \cdot(\vec{b} \times \vec{c})=(2 \hat{i}+\hat{j}+3 \hat{k}) \cdot(3 \hat{i}+5 \hat{j}-7 \hat{k})$
$=2 \times 3+1 \times 5+3 \times(-7)$
$=6+5-21=-10$
View full question & answer→MCQ 2201 Mark
If $|\vec{a}-\vec{b}|=|\vec{a}|=|\vec{b}|=1$, then the angle between $\vec{a}$ and $\vec{b}$ is
- ✓
$\frac{\pi}{3}$
- B
$\frac{3 \pi}{4}$
- C
$\frac{\pi}{2}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: A. $\frac{\pi}{3}$
(a) : Given, $|\vec{a}-\vec{b}|=|\vec{a}|=|\vec{b}|=1$
$
\Rightarrow|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} \Rightarrow 1=1+1-2|\vec{a}||\vec{b}| \cos \theta
$
(Here $\theta$ is angle between $\vec{a}$ and $\vec{b}$ )
$
\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
$
View full question & answer→MCQ 2211 Mark
If the angle between $\hat{i}+\hat{k}$ and $\hat{i}+\hat{j}+a \hat{k}$ is $\frac{\pi}{3}$ then the value of $a$ is
- A
$2$
- B
$-4$ or $0$
- ✓
$0$
- D
$2$ or $-2$
AnswerWe have, $\cos \frac{\pi}{3}=\frac{(\hat{i}+\hat{k}) \cdot(\hat{i}+\hat{j}+a \hat{k})}{\sqrt{2} \sqrt{1+1+a^2}}$
$\Rightarrow \frac{1}{2}=\frac{1+a}{\sqrt{2} \sqrt{2+a^2}}$
$\Rightarrow \frac{1}{4}=\frac{(1+a)^2}{2\left(2+a^2\right)} ...(i)$
$\Rightarrow 2+a^2=2\left(1+a^2+2 a\right)$
$\Rightarrow a^2+4 a=0$
$\Rightarrow a=0,-4$
But $a=-4$ does not satisfy equation $(i),$
Hence, the required value of $a$ is $0$.
View full question & answer→MCQ 2221 Mark
Which of the following is the vector in the direction of the vector $\hat{i}-2 \hat{j}+2 \hat{k}$ that has magnitude $9\ ?$
- A
$\hat{i}-2 \hat{j}+2 \hat{k}$
- B
$\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}$
- ✓
$3(\hat{i}-2 \hat{j}+2 \hat{k})$
- D
$9(\hat{i}-2 \hat{j}+2 \hat{k})$
AnswerCorrect option: C. $3(\hat{i}-2 \hat{j}+2 \hat{k})$
Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$
$\therefore|\vec{a}|=\sqrt{1+4+4}=\sqrt{9}=3$
$\therefore$ Required vector $=\frac{9(\hat{i}-2 \hat{j}+2 \hat{k})}{3}=3(\hat{i}-2 \hat{j}+2 \hat{k})$
View full question & answer→MCQ 2231 Mark
If $A$ and $B$ are the points $(-3,4,-8)$ and $(5,-6,4)$ respectively, then find the ratio in which $y z$-plane divides $\overrightarrow{A B}$.
- A
$5: 2$
- B
$7: 5$
- ✓
$3: 5$
- D
$5: 3$
AnswerCorrect option: C. $3: 5$
(c) : Let $\vec{a}=-3 \hat{i}+4 \hat{j}-8 \hat{k}, \vec{b}=5 \hat{i}-6 \hat{j}+4 \hat{k}$
Let $C(\vec{c})$ be the point in $y z$-plane which divides $\overrightarrow{A B}$ in the ratio $r: 1$ internally.
Then, $0=\frac{5 r-3}{r+1} \quad(\because$ In $y z$-plane, $x=0)$
$\Rightarrow \quad 5 r-3=0 \Rightarrow r=\frac{3}{5}$
Thus required ratio is $3: 5$
View full question & answer→MCQ 2241 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
$\frac{5}{\sqrt{3}}$
AnswerCorrect 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 $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{10}{\sqrt{6}}$.
View full question & answer→MCQ 2251 Mark
If $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=4 \hat{i}+4 \hat{j}-2 \hat{k}$ then find the angle between the vectors $\vec{a}$ and $\vec{b}$.
- A
- B
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{2}$
(c) : We have, $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=4 \hat{i}+4 \hat{j}-2 \hat{k}$
Now, $\vec{a} \cdot \vec{b}=(2 \hat{i}-\hat{j}+2 \hat{k}) \cdot(4 \hat{i}+4 \hat{j}-2 \hat{k})$
$=8-4-4=0$. Therefore, $\vec{a} \cdot \vec{b}=0$
$
\Rightarrow \cos \theta=0
$
So, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.
View full question & answer→MCQ 2261 Mark
If $A B C D$ is a rhombus, whose diagonals intersect at $E$, then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals
- ✓
$\overrightarrow{0}$
- B
$\overrightarrow{A D}$
- C
$2 \overrightarrow{B C}$
- D
$2 \overrightarrow{A D}$
AnswerCorrect option: A. $\overrightarrow{0}$
(a) : $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$
$
=\overrightarrow{E A}+\overrightarrow{E B}-\overrightarrow{E A}-\overrightarrow{E B}
$ [As diagonals of a rhombus bisect each other]
$=\overrightarrow{0}$
View full question & answer→MCQ 2271 Mark
$(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2$ is equal to
- A
- B
$|\vec{a}|$
- C
$-\vec{a}$
- ✓
$|\vec{a}|^2$
AnswerCorrect option: D. $|\vec{a}|^2$
(d) : Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k} \Rightarrow(\vec{a} \cdot \hat{i})^2=x^2$
Similarly, $(\vec{a} \cdot \hat{j})^2=y^2$ and $(\vec{a} \cdot \hat{k})^2=z^2$
$
\therefore \quad(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2=x^2+y^2+z^2=|\vec{a}|^2
$
View full question & answer→MCQ 2281 Mark
Which of the following is the magnitude of the vector $\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ ?
Answer(b) : Here, $\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$
$\therefore \quad$ Its magnitude $=|\vec{a}|$
$
=\sqrt{3^2+(-2)^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7
$
View full question & answer→MCQ 2291 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 $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.
AnswerCorrect option: A. $-4 \hat{j}-\hat{k}$
The given vectors are
$\vec{a}=\hat{i}-2 \hat{j}+\hat{k},$
$\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k},$
$\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$
$\therefore$ Required sum $=\vec{a}+\vec{b}+\vec{c}$
$=(\hat{i}-2 \hat{j}+\hat{k})+(-2 \hat{i}+4 \hat{j}+5 \hat{k})+(\hat{i}-6 \hat{j}-7 \hat{k})$
$=-4 \hat{j}-\hat{k}$
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