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MATHS M.C.Q (1 Marks)

Vector Algebra · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 65 questions.

Questions · Page 2 of 2

M.C.Q (1 Marks)

Question 511 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
Answer
$(b) :$ Given, $|\vec{a}|=|\vec{b}|, \theta=60^{\circ}$ and $\vec{a} \cdot \vec{b}=\frac{9}{2}$
Now, $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\Rightarrow \cos 60^{\circ}=\frac{9 / 2}{|\vec{a}|^2} \Rightarrow \frac{1}{2}=\frac{9 / 2}{|\vec{a}|^2}$
$\Rightarrow|\vec{a}|^2=9 \Rightarrow|\vec{a}|=3 \therefore|\vec{a}|=|\vec{b}|=3$
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Question 521 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
$(d) : |\vec{a}|=10,|\vec{b}|=2, \vec{a} \cdot \vec{b}=12$
We know that, $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
$\Rightarrow 12=10 \times 2 \cos \theta \Rightarrow \cos \theta=\frac{3}{5}$
$\therefore \sin \theta=\frac{4}{5}$
Now, $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=10 \times 2 \times \frac{4}{5}=16$
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Question 531 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
Answer
$(b) :$ Clearly, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.
$\Rightarrow \vec{a} \cdot \vec{b}=0$
$\therefore|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1+1+0=2$
$\Rightarrow|\vec{a}+\vec{b}|=\sqrt{2}$
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Question 541 Mark
Find the value of $\lambda$ for which the vectors $3 \hat{i}-6 \hat{j}+\hat{k}$ and $2 \hat{i}-4 \hat{j}+\lambda \hat{k}$ are parallel.
Answer
(a) : $\vec{a}=3 \hat{i}-6 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}-4 \hat{j}+\lambda \hat{k}$
Since, $\vec{a}$ and $\vec{b}$ are parallel $\quad \therefore \quad \vec{a} \times \vec{b}=\overrightarrow{0}$
$
\begin{aligned}
\Rightarrow & \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -6 & 1 \\
2 & -4 & \lambda
\end{array}\right|=\overrightarrow{0} \\
\Rightarrow & (-6 \lambda+4) \hat{i}-(3 \lambda-2) \hat{j}+(-12+12) \hat{k}=\overrightarrow{0} \\
\Rightarrow & (-6 \lambda+4) \hat{i}+(2-3 \lambda) \hat{j}=0 \hat{i}+0 \hat{j}
\end{aligned}
$
Comparing coefficients of $\hat{i}$ and $\hat{j}$, we get $-6 \lambda+4=0$ and $2-3 \lambda=0 \Rightarrow \lambda=2 / 3$
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Question 551 Mark
Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$, $\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.
Answer
$(a)$: The given vectors are
$\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}, \vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$
$\therefore \text { Required sum }=\vec{a}+\vec{b}+\vec{c}$
$=(\hat{i}-2 \hat{j}+\hat{k})+(-2 \hat{i}+4 \hat{j}+5 \hat{k})+(\hat{i}-6 \hat{j}-7 \hat{k})$
$=-4 \hat{j}-\hat{k}$
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Question 561 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?
$(i) \theta=\frac{\pi}{2}$
$(ii) \theta < \frac{\pi}{3}$
$(iii) \pi \geq \theta > \frac{2 \pi}{3}$
$(iv) \cos \theta < -\frac{1}{2}$
Answer
$\text { (c) : }|\vec{a}+\vec{b}| <1$
$\Rightarrow|\vec{a}+\vec{b}|^2<1$
$\Rightarrow|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b} <1$
$\Rightarrow 1+1+2 \vec{a} \cdot \vec{b} <1 \quad[\because|\vec{a}|=|\vec{b}|=1]$
$\Rightarrow \vec{a} \cdot \vec{b}<-\frac{1}{2}$
$\Rightarrow|\vec{a}||\vec{b}| \cos \theta<-\frac{1}{2}$
$\Rightarrow 1 \times 1 \times \cos \theta<-\frac{1}{2}$
$\Rightarrow \cos \theta<-\frac{1}{2}$
$\Rightarrow-1 \leq \cos \theta<-\frac{1}{2}$
$\Rightarrow \pi \geq \theta>\frac{2 \pi}{3}$
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Question 571 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 $(\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} \cdot(\vec{b} \times \vec{c})$.
Answer
(c) : Let $\vec{b}=\hat{i}$ and $\vec{c}=\hat{j}$ and $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ Now, $\vec{a} \cdot \vec{b}=a_1, \vec{a} \cdot \vec{c}=a_2$ and $\vec{a} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}=\vec{a} \cdot \hat{k}=a_3$
$
\begin{aligned}
\therefore \quad & (\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|}(\vec{b} \times \vec{c}) \\
\quad & =a_1 \vec{b}+a_2 \vec{c}+a_3 \hat{k}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}=\vec{a}
\end{aligned}
$
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Question 581 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
Answer
$(c) :$ We have $\vec{a}, \vec{b}, \vec{c}$ are unit vectors.
Therefore, $|\vec{a}|=1,|\vec{b}|=1$ and $|\vec{c}|=1$
Also, $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ (given)
$\Rightarrow \quad|\vec{a}+\vec{b}+\vec{c}|^2=0$
$\Rightarrow \quad|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow 3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow \quad(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}$
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Question 591 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?
Answer
(b) : Given,
$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & p & q\end{array}\right|=0$
$
\begin{array}{ll}
\Rightarrow & \hat{i}(6 q-27 p)-\hat{j}(2 q-27)+\hat{k}(2 p-6)=0 \\
\Rightarrow & 6 q-27 p=0,2 q-27=0 \text { and } 2 p-6=0 \\
\Rightarrow & q=\frac{27}{2} \text { and } p=3 .
\end{array}
$
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Question 601 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
Answer
$(b) :$ We have, $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}, \vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$
$\therefore 3 \vec{a}+2 \vec{b}=3(\hat{i}+\hat{j}-2 \hat{k})+2(2 \hat{i}-4 \hat{j}+5 \hat{k})$
$=(3 \hat{i}+3 \hat{j}-6 \hat{k})+(4 \hat{i}-8 \hat{j}+10 \hat{k})=7 \hat{i}-5 \hat{j}+4 \hat{k}$
$\therefore \quad$ The direction ratios of the vector $3 \vec{a}+2 \vec{b}$ are $7,-5,4$.
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Question 611 Mark
Which of these are the direction cosines of the vector $-2 \hat{i}+\hat{j}-5 \hat{k}$ ?
Answer
(b): We have, $\vec{a}=-2 \hat{i}+\hat{j}-5 \hat{k}$
Direction cosines of the given vector are
$
\left.\begin{array}{rl}
\left(\frac{-2}{\sqrt{(-2)^2+(1)^2+(-5)^2}}, \frac{1}{\sqrt{(-2)^2+(1)^2+(-5)^2}}\right. ,\frac{-5}{\sqrt{(-2)^2+(1)^2+(-5)^2}}
\end{array}\right)
$
$
=\left(\frac{-2}{\sqrt{4+1+25}}, \frac{1}{\sqrt{4+1+25}}, \frac{-5}{\sqrt{4+1+25}}\right)
$
$\therefore$ Direction cosines are $\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)$
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Question 621 Mark
$(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2$ is equal to
Answer
(d) : Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k} \Rightarrow(\vec{a} \cdot \hat{i})^2=x^2$
Similarly, $(\vec{a} \cdot \hat{j})^2=y^2$ and $(\vec{a} \cdot \hat{k})^2=z^2$
$
\therefore \quad(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2=x^2+y^2+z^2=|\vec{a}|^2
$
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MCQ 631 Mark
If $\vec{a}=\overparen{i}-7 \overparen{j}+7 \overparen{k}$ and $\vec{b}=3 \overparen{i}-2 \overparen{j}+2 \overparen{k}$ then value of $|\vec{a} \times \vec{b}|$ is -
  • A
    $19 \sqrt{2}$
  • ✓
    $2 \sqrt{19}$
  • C
    19
  • D
    None of these
Answer
Correct option: B.
$2 \sqrt{19}$
B
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MCQ 641 Mark
If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$ then $|\vec{x}|$ is -
  • A
    $2 \sqrt{2}$
  • B
    $-9$
  • ✓
    3
  • D
    9
Answer
Correct option: C.
3
C
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MCQ 651 Mark
Value of $\widehat{i} \cdot(\widehat{j} \times \widehat{k})+\widehat{j} \cdot(\widehat{i} \times \widehat{k})+\widehat{k} \cdot(\widehat{i} \times \widehat{j})$ is -
  • ✓
    1
  • B
    $-1$
  • C
    3
  • D
    $0$
Answer
Correct option: A.
1
A
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M.C.Q (1 Marks) - Page 2 - MATHS STD 12 Science Questions - Vidyadip