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MATHS M.C.Q (1 Marks)

VECTOR ALGEBRA · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 172 questions.

Questions · Page 2 of 4

M.C.Q (1 Marks)

Question 511 Mark
If the position vectors of P, Q are respectively 5a + 4b and 3a - 2b then $\vec{\text{QP​}}=$
  1. 2a + 6b
  2. 2a − 6b
  3. 2a + 5b
  4. 2a − 5b
Answer
  1. 2a + 6b
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Question 521 Mark
Choose the correct answer from the given four options.
The vector in the direction of the vector $\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ that has magnitude 9 is:
  1. $\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
  2. $\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{3}$
  3. $3(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
  4. $9(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
Answer
  1. $3(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
Solution:
Let $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
Any vector in the direction of a vector is given by $\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{1^2+2^2+2^2}}=\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{3}$
$\therefore$ Vector in the direction of with magnitude 9 is $9\cdot\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=9\cdot\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{3}$
$=3(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
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Question 531 Mark
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors inclined at an angle $\theta,$ then the value of $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ is:
  1.  $2\sin\frac{\theta}{2}$
  2. $2\sin\theta$
  3. $2\cos\frac{\theta}{2}$
  4. $2\cos\theta$
Answer
  1. $2\sin\frac{\theta}{2}$
Solution:
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$=1\times1\cos\theta$ (Because $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors)
$=\cos\theta\dots(1)$
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
 $=1+1-2\cos\theta$ [using (1)]
$=2-2\cos\theta$
$=2(1-\cos)$
$=2\big(2\sin^2\frac{\theta}{2}\big)$
$=4\sin^2\frac{\theta}{2}$
$\therefore\big|\vec{\text{a}}-\vec{\text{b}}\big|=2\sin\frac{\theta}{2}$
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Question 541 Mark
If u, v, w are non-coplanar vector and p, q are real numbers, then the equality [3u pv pw] - [pv w qw] - [2w qv qu] = 0 holds for:
  1.  Exactly two values of (p, q)
  2. More than but not all values of (p, q)
  3. All values of (p, q)
  4. Exactly one values of (p, q) 
Answer
  1. Exactly two values of (p, q)
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Question 551 Mark
If $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|,$ then $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=$
  1. Positive
  2. Negetive
  3. 0
  4. None of these
Answer
  1. 0
Solution:
Given that
$|\vec{\text{a}}|=|\vec{\text{a}}|$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$|\vec{\text{a}}^2-|\vec{\text{a}}|^2$
$=0$
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Question 561 Mark
The position vectors of P and Q are respectively a and b.If R is a point on PQ, PQ such that PR = 5PQ, then the position vector of R is:
  1. 5b − 4a
  2. 5b + 4a
  3. 4b − 5a
  4. 4b + 5a
Answer
  1. 5b − 4a
Solution:
Given condition 
PR = 5PQ
R − P = 5(Q − P)
R = 5Q − 5P + P
R = 5Q − 4P
R = 5b − 4a
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Question 571 Mark
If $\vec{\text{a}}$ be the position vector whose tip is (5, -3) find the coordinates of a point B such that $\vec{\text{AB}}=\vec{\text{a}}$ the coordinates of A being (4, -1):
  1. (9, -4)
  2. (-9, -4)
  3. (9, 4)
  4. None of these
Answer
  1. (9, -4)
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Question 581 Mark
Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ be three zero vectors such that $\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$ If the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6},$ then $\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$ is equal to:
  1.  $0$
  2. $1$
  3. $\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
  4. $\frac{3}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
Answer
  1. $\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
Solution:
We have
$\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$
$=\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big]^2$ (By definition of scalar triple product)
$=\big[\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|\cos0^\circ\big]^2$ $\big(\therefore\vec{\text{a}}\times\vec{\text{b}}$ is parallel to vector $\vec{\text{c}}$ as $\vec{\text{c}}$ is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}\big)$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\frac{\pi}{6}\big)^2$ $\big(\therefore\big|\vec{\text{c}}\big|=1$ and angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}\big)$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\frac{1}{2}\big)^2$
$=\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
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Question 591 Mark
Choose the correct answer from the given four options.If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three vectors such that $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0}$ and $|\vec{\text{a}}|=2,|\vec{\text{b}}|=3$ and $|\vec{\text{c}}|=5,$ 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. 0.
  2. 1.
  3. -19.
  4. 38.
Answer
  1. -19.
Solution:
Here, $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0}$ and $\vec{\text{a}}^2=4,\vec{\text{b}}^2=9,\vec{\text{c}}^2=25$
$\therefore(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})\cdot(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})=0$
$\Rightarrow\vec{\text{a}}^2+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}^2+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{b}}+\vec{\text{c}}^2=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$ $[\because\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}}]$
$\Rightarrow4+9+25+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$
$\Rightarrow\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}=0$
$\Rightarrow\frac{-38}{2}=-19$
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Question 601 Mark
If $ \overrightarrow {\text{ a }}$ is vector of magnitude x, m is non-zero scalar and $\text{m}\overrightarrow { \text{a} }$ is a unit vector then x in terms of m is:
  1. $\text{m}=\text{x}$
  2. $\text{x}=\mid{\text{m}}\mid$
  3. $\text{x}=\frac{1}{\mid\text{m}\mid}$
  4. $\text{x}=\frac{\text{m}}{2}$
Answer
  1. $\text{x}=\frac{1}{\mid\text{m}\mid}$
Solution:
Given, $\mid\text{m}\vec{\text{a}}\mid=1$
$\Rightarrow\mid\text{m}\mid\mid\vec{\text{a}}\mid=1$
$\Rightarrow\mid\text{m}\mid\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{\mid\text{m}\mid}$
Remember, modulus can never be negative.
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Question 611 Mark
If in a $\triangle\text{ABC}$, $\text{A}=(0,0),\ \text{B}=(3,3\sqrt3),\ \text{C}=(-3\sqrt3,3)$, then the vecctor of magnitude $2\sqrt2$ units directed along AO, where O is the circumcenter of $\triangle\text{ABC}$ is,
  1. $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
  2. $(1+\sqrt3)\hat{\text{i}}+(1-\sqrt3)\hat{\text{j}}$
  3. ​​​​​​​​​​​​​​$(1+\sqrt3)\hat{\text{i}}+(\sqrt3-1)\hat{\text{j}}$
  4. None of these
Answer
  1. $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
Solution:

$\Big|\overrightarrow{\text{AO}}\Big|=2\sqrt2$
$\Big|\overrightarrow{\text{AO}}\Big|=\Big|\overrightarrow{\text{BO}}\Big|=\Big|\overrightarrow{\text{CO}}\Big|=2\sqrt2=\text{R}$
Let the position vector of O be $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$
$\Big|\overrightarrow{\text{AO}}\Big|=\sqrt{\text{x}^2+\text{y}^2}$
$\therefore\ \text{x}^2+\text{y}^2=8\ \dots(1)$
Also, $\Big|\overrightarrow{\text{BO}}\Big|=\Big|\overrightarrow{\text{CO}}\Big|$
$\sqrt{(\text{x}-3)^2+(\text{y}-3\sqrt3)^2}\\=\sqrt{(\text{x}+3\sqrt3)^2+(\text{y}-3)^2}$
$\text{x}^2-6\text{x}+9+\text{y}^2-6\sqrt3\text{y}+27\\=\text{x}^2+6\sqrt3\text{x}+27+\text{y}^2-6\text{y}+9$
$\text{y}(6-6\sqrt3)=\text{x}(6\sqrt3+6)$
$\text{y}=\frac{\text{x}(1+\sqrt3)}{(1-\sqrt3)}\ \dots(2)$
Substituting y from (2) in (1) we get,
$(1-\sqrt3)^2\text{x}^2+(1+\sqrt3)^2\text{x}^2=8(1-\sqrt3)^2$
$\text{x}^2\times8=8(1-\sqrt3)^2$
$\text{x}=1-\sqrt3$
$\text{y}=1+\sqrt3$
$\therefore$ The position vector of O is $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
$\overrightarrow{\text{AO}}=(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
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Question 621 Mark
If the angle between the vectors $\text{x}\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}$ and $\text{x}\hat{\text{i}}-\text{x}\hat{\text{j}}+4\hat{\text{k}}$ is acute, then x lies in the interval:
  1. (-4, 7)
  2. [-4, 7]
  3. R - [-4, 7]
  4. R - (4, 7)
Answer
  1. R - [-4, 7]
Solution:
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{\text{x}^2-3\text{x}-28}{\sqrt{\text{x}^2+3^2+49}\sqrt{\text{x}^2+\text{x}^2+4^2}}$
For $\theta$ to be acute,
$\cos\theta>0$
$\Rightarrow\text{x}^2-3\text{x}-28>0$
$\Rightarrow(\text{x}-7)(\text{x}+4)>0$
$\Rightarrow\text{x}\in(-\infty,-4)\cup(7,\infty)$
$\Rightarrow\text{x}\in\text{R}-[-4,7]$
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Question 631 Mark
The orthogonal projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is:
  1. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{a}}}{|\vec{\text{a}}|^2}$
  2. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
  3. $\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
  4. $\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Answer
  1. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
Solution:
The orthogonal projection of $​\vec{\text{a}}$ on $​\vec{\text{b}}$ is
$\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
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Question 641 Mark
Namita walks 14 metres towards west, then turns to her right and walks 14 metres and then turns to her left and walks 10 metres.Again turning to her left she walks 14 metres.What is the shortest distance (in metres) between her starting point and the present position?
  1. 10
  2. 24
  3. 28
  4. 38
Answer
  1. 24
Solution:
So, shortest distance = 24
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Question 651 Mark
Four persons P, Q, R and S are initially at the four corners of a square side d. Each person now moves with a constant speed v in such a way that P always moves directly towards Q, Q towards R, R towards S, and S towards P. The four persons will meet after time.
  1. $\frac{\text{d}}{2\text{v}}$
  2. $\frac{\text{d}}{\text{v}}$
  3. $\frac{\text{3d}}{2\text{v}}$
  4. They will never meet
Answer
  1. $\frac{\text{d}}{\text{v}}$
Solution:
Here, velocity components will be vcos $45=\frac{\text{v}}{\sqrt{2}}$
And, displacement will be $\frac{\text{d}}{\sqrt{2}}$
So time taken will be
$\text{t}=\frac{\text{d}}{\text{v}}$
$=\frac{\frac{\text{d}}{\sqrt{2}}}{\frac{\text{v}}{\sqrt{2}}}=\frac{\text{d}}{\text{v}}$
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MCQ 661 Mark
If the vectors $4\hat{\text{i}}+11\hat{\text{j}}+\text{m}\hat{\text{k}},7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ are coplanar, then $m =$
  • A
    $0$
  • B
    $38$
  • C
    $-10$
  • ✓
    $10$
Answer
Correct option: D.
$10$
Let
$\vec{\text{a}}=4\hat{\text{i}}+11\hat{\text{j}}+{\text{m}}\hat{\text{k}}$
$\vec{\text{b}}=7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are coplanar if their scalar triple product is zero, i.e. $\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}4&11&\text{m}\\7&2&6\\1&5&4 \end{vmatrix}=0$
$\Rightarrow 4(8-30)-11(28-6)+\text{m}(35-2)=0$
$\Rightarrow-88-242+33\text{m}=0$
$\Rightarrow33\text{m}=330$
$\therefore\text{m}=10$
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Question 671 Mark
If $\vec{\text{x}}$ is a vector in the direction of (2, -2, 1) of magnitude 6 and $\vec{\text{y}}$ is a vector in the direction of  (1, 1, -1) of magnitude $\sqrt{3}$ then $\mid\vec{\text{x}}+2\vec{\text{y}}\mid=$
  1. $40$
  2. $\sqrt{35}$
  3. $\sqrt{17}$
  4. $2\sqrt{10}$
Answer
  1. $2\sqrt{10}$
Solution:
They given x directionwe need to find unit vector in that direction and multiply with the magnitude of x they given y directionwe need to find unit vector in that direction and multiply with the magnitude of $\vec{\text{x}}\text{y}$
$\frac{{6}\Big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\Big)}{3},\vec{\text{y}}=\frac{{\sqrt{3}}\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)}{\sqrt{3}},$
so $\mid\vec{\text{x}}+2\vec{\text{y}}\mid=\mid6\hat{\text{i}}-2\hat{\text{j}}\mid=\sqrt{40}=2\sqrt{10}$
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Question 681 Mark
The vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ satisfy the equation $2\vec{\text{a}}+\vec{\text{b}}=\vec{\text{p}}$ and $\vec{\text{a}}+2\vec{\text{b}}=\vec{\text{q}},$ where $\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{q}}=\hat{\text{i}}-\hat{\text{j}}.$ If $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}},$ then:
  1. $\cos \theta = \frac{4}{5}$
  2. $\sin \theta = \frac{1}{\sqrt{2}}$
  3. $\cos \theta = -\frac{4}{5}$
  4. $\cos \theta = -\frac{3}{5}$
Answer
  1. $\cos \theta = -\frac{4}{5}$
Solution:
Given that
$2\vec{\text{a}}+\vec{\text{b}}=\vec{\text{p}}\dots(1)$
$\vec{\text{a}}+2\vec{\text{b}}=\vec{\text{q}}\dots(2)$
Solving these two we get
$\vec{\text{a}}=\frac{2\vec{\text{p}}-\vec{\text{q}}}{3},\vec{\text{b}}=\frac{2\vec{\text{q}}-\vec{\text{p}}}{3}$
And we have
$\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{q}}=\hat{\text{i}}-\hat{\text{j}}$
Substituting the values of $\vec{\text{p}}$ and $\vec{\text{q}},$ we get
$\vec{\text{a}}=\frac{2\vec{\text{p}}-\vec{\text{q}}}{3}=\frac{2\big(\hat{\text{i}}+\hat{\text{j}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)}{3}=\frac{\hat{\text{i}}+3\hat{\text{j}}}{3}$
$\Rightarrow|\vec{\text{a}}|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3}$
$\vec{\text{b}}=\frac{2\vec{\text{q}}-\vec{\text{p}}}{3}=\frac{2\big(\hat{\text{i}}-\hat{\text{j}}\big)-\big(\hat{\text{i}}+\hat{\text{j}}\big)}{3}=\frac{\hat{\text{i}}-3\hat{\text{j}}}{3}$
$\Rightarrow\big|\vec{\text{b}}\big|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3}$
$\vec{\text{a}}.\vec{\text{b}}=\frac{1}{9}(1-9)=\frac{-8}{9}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\frac{-8}{9}=\frac{\sqrt{10}}{3}\times\frac{\sqrt{10}}{3}\cos\theta$
$\Rightarrow\frac{-8}{9}=\frac{10}{9}\cos\theta$
$\Rightarrow\cos\theta=\frac{-8}{9}\times\frac{9}{10}=\frac{-4}{5}$
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Question 691 Mark
The scalar product of 5i + j - 3k and 3i - 4j + 7k is:
  1. 15
  2. -15
  3. 10
  4. -10
Answer
  1. -10
Solution:
Let A = 5i + j – 3k
B = 3i – 4j + 7k
A.B = (5i + j - 3k) (3i - 4j + 7k)
= 5.3 + 1.(-4) + (-3).7
= 15 - 4 - 21
= -10
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Question 701 Mark
The value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is:
  1. $|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
  2. $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
  3. $|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\big(\vec{\text{a}}.\vec{\text{b}}\big)$
  4. $|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-\vec{\text{a}}.\vec{\text{b}}$
Answer
  1. $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
Solution:
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(\cos^2\theta+\sin^2\theta)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1)
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
Thus, the value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2.$
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Question 711 Mark
If $\vec{\text{a}},\vec{\text{b}}$ represent the diagonals of a rhombus, then:
  1. $\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$
  2. $\vec{\text{a}}.\vec{\text{b}}=0$
  3. $\vec{\text{a}}.\vec{\text{b}}=1$
  4. $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}$
Answer
  1. $\vec{\text{a}}.\vec{\text{b}}=0$
Solution:
We know that the diagonals in a rhombus $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular.
Therefore, their dot product is zero.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
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Question 721 Mark
In figure, which of the following is not true?
  1. $​​​​​\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec0$
  2. $​​​​​\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec0$
  3. $​​​​​\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec0$
  4. $​​​​​\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec0$
Answer
  1. $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec0$
Solution:
We have, LHS = $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\overrightarrow{\text{AC}}-\overrightarrow{\text{CA}}$ $\Big[\because​​\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=-\overrightarrow{\text{CA}}-\overrightarrow{\text{CA}}$
$=-2\overrightarrow{\text{CA}}$
So, $\text{LHS}\neq\text{RHS}$
Hence, It is not true.
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Question 731 Mark
If $\vec{\text{a}},\ \vec{\text{b}}$ are the vectors forming consecutive sides of a regular hexagon ABCDEF, then the vector representing side CD is,
  1. $\vec{\text{a}}+\vec{\text{b}}$
  2. $\vec{\text{a}}-\vec{\text{b}}$
  3. $\vec{\text{b}}-\vec{\text{a}}$
  4. $-\big(\vec{\text{a}}+\vec{\text{b}}\big)$
Answer
  1. $\vec{\text{b}}-\vec{\text{a}}$
Solution:
Let ABCDEF be a regular hexagon such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{BC}}=\vec{\text{b}}$. We know, AD is parallel to BC such that AD = 2BC.
$\therefore\ \overrightarrow{\text{AD}}=2\overrightarrow{\text{BC}}=2\vec{\text{b}}$
In $\triangle{\text{ABC}}$, we have
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\Rightarrow\ \vec{\text{a}}+\vec{\text{b}}=\overrightarrow{\text{AC}}$
In $\triangle{\text{ACD}}$, we have
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$
$\Rightarrow\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}-\overrightarrow{\text{AC}}$
$\Rightarrow\overrightarrow{\text{CD}}=2\vec{\text{b}}-\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$\Rightarrow\overrightarrow{\text{CD}}=\vec{\text{b}}-\vec{\text{a}}$
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Question 741 Mark
Two or more vectors having the same initial point are:
  1. Coinitial vectors
  2. colinear vectors
  3. equal vectors
  4. Cannot say
Answer
  1. Coinitial vectors
Solution:
Two or more vectors having same initial points are known as co-initial vectors.
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Question 751 Mark
A person travels 12km in the southward direction and then travels 5km to the right and then travels 15km toward the right and finally travels 5km towards the east, how far is he from his starting place?
  1. 5.5kms
  2. 3km
  3. 13km
  4. 6.4km
Answer
  1. 3km
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Question 761 Mark
If A(6, 3, 2), B(5, 1, 4), C(3, −4, 7), D(0, 2, 5) are four points, then projection of CD on AB is:
  1. $-\frac{13}{7}$
  2. $-\frac{13}{7}$
  3. $-\frac{3}{13}$
  4. $-\frac{7}{13}$
Answer
  1. $-\frac{13}{7}$
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Question 771 Mark
The vector equation of the plane passing through $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},$ is $\vec{\text{r}}=\alpha\vec{\text{a}}+\beta\vec{\text{b}}+\gamma\vec{\text{c}}$, provided that,
  1. $\alpha+\beta+\gamma=0$
  2. $\alpha+\beta+\gamma=1$
  3. $\alpha+\beta=\gamma$
  4. $\alpha^2+\beta^2+\gamma^2=1$
Answer
  1. $\alpha+\beta+\gamma=1$
Solution:
Given: A plane passing through $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$.
⇒ Lines $\vec{\text{a}}-\vec{\text{b}}$ and $\vec{\text{c}}-\vec{\text{a}}$ lie on the plane.
The parmetric equation of the plane can be written as:
$\vec{\text{r}}=\vec{\text{a}}+\lambda_1\big(\vec{\text{a}}-\vec{\text{b}}\big)+\lambda_2\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$\vec{\text{r}}=\vec{\text{a}}(1+\lambda_1+\lambda_2)-\lambda_1\vec{\text{b}}+\lambda_2\vec{\text{c}}$
Given that $\vec{\text{r}}=\alpha\vec{\text{a}}+\beta\vec{\text{b}}+\gamma\vec{\text{c}}$
$\therefore\alpha+\beta+\gamma=1+\lambda_1-\lambda_2-\lambda_1+\lambda_2$
$\alpha+\beta+\gamma=1$
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Question 781 Mark
The system of vectors i, j, k is:
  1. Orthogonal
  2. Collinear
  3. Coplana
  4. None of these
Answer
  1. 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 791 Mark
The vector $\cos\alpha\cos\beta\hat{\text{i}}+\cos\alpha\sin\beta\hat{\text{j}}+\sin\alpha\hat{\text{k}}$ is a,
  1. Null vector.
  2. Unit vector.
  3. Constant vector.
  4. None of these.
Answer
  1. Unit vector
Solution:
Given: The vector $\cos\alpha\cos\beta\hat{\text{i}}+\cos\alpha\sin\beta\hat{\text{j}}+\sin\alpha\hat{\text{k}}$. Then,
$\big|\cos\alpha\cos\beta\hat{\text{i}}+\cos\alpha\sin\beta\hat{\text{j}}+\sin\alpha\hat{\text{k}}\big|$
$=\sqrt{\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha}$
$=\sqrt{\cos^2\alpha+\sin^2\alpha}=1$
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Question 801 Mark
Choose the correct answer from the given four options.
The vectors from origin to the points A and B are $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}},$ respectively, then the area of the triangle OAB is:
  1. $340$
  2. $\sqrt{25}$
  3. $\sqrt{229}$
  4. $\frac{1}{2}\sqrt{229}$
Answer
  1. $\frac{1}{2}\sqrt{229}$
Solution:
$\therefore$ Area of $\triangle\text{OAB}=\frac{1}{2}|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}|$
$=\frac{1}{2}|(2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})\times(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})|$
$=\frac{1}{2}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-3&2\\2&3&1 \end{vmatrix}$
$=\frac{1}{2}|[\hat{\text{i}}(-3-6)-\hat{\text{j}}(2-4)+\hat{\text{k}}(6+6)]|$
$=\frac{1}{2}|-9\hat{\text{i}}+2\hat{\text{j}}+12\hat{\text{k}}|$
$\therefore$ Area of $\triangle\text{OAB}=\frac{1}{2}\sqrt{81+4+144}$
$=\frac{1}{2}\sqrt{229}$
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Question 811 Mark
Point (4, 0) lies on:
  1. $\vec{\text{XO}}$
  2. $\vec{\text{YO}}$
  3. $\vec{\text{OX}}$
  4. $\vec{\text{OY}}$
Answer
  1. $\vec{\text{OX}}$
Solution:
$\vec{\text{XO}}$ is positive x-axis, so (4, 0) lies on it.
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Question 821 Mark
Choose the correct answer from the given four options.
The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is:
  1. $-\hat{\text{i}}+12\hat{\text{j}}+4\hat{\text{k}}$
  2. $-5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
  3. $-5\hat{\text{i}}+12\hat{\text{j}}+4\hat{\text{k}}$
  4. $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Answer
  1. $-5\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
Solution:
Given points are (2, 5, 0) and (–3, 7, 4).
Thus, the required vector $=(-3-2)\hat{\text{i}}+(7-5)\hat{\text{j}}+(4-0)\hat{\text{k}}$
$=-5\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
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Question 831 Mark
What is the length of the longer diagonal of the parallelogram constructed on $5\vec{\text{a}}+2\vec{\text{b}}$ and $\vec{\text{a}}-3\vec{\text{b}}$ if it is given that $|\vec{\text{a}}|=2\sqrt{2},\big|\vec{\text{b}}\big|=3$ and the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{4}$?
  1. $15$
  2. $\sqrt{113}$
  3. $\sqrt{593}$
  4. $\sqrt{369}$
Answer
  1. $\sqrt{593}$
Solution:
Let ABCD be a parallelogram in which
side $\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}=5\vec{\text{a}}+2\vec{\text{b}}$
and $\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}=\vec{\text{a}}-3\vec{\text{b}}$
and diagonals are AC and BD.
Now, $\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}$
$=\big(5\vec{\text{a}}+2\vec{\text{b}\big)}+\big(\vec{\text{a}}-3\vec{\text{b}}\big)$
$=6\vec{\text{a}}-\vec{\text{b}}$
$\therefore\big|\overrightarrow{\text{AC}}\big|=\big|6\vec{\text{a}}-\vec{\text{b}\big|}$
$=\sqrt{|6\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\times|6\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\theta}$
$=\sqrt{36|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-12\times|\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\frac{\pi}{4}}$
$=\sqrt{36|2\sqrt{2}|^2+|3|^2-12\times|2\sqrt{2}|\times|3|\times\frac{1}{\sqrt{2}}}$
$=\sqrt{288+9-72}$
$=\sqrt{225}=15\text{ units}$
$\overrightarrow{\text{BD}}=\overrightarrow{\text{BA}}+\overrightarrow{\text{BD}}$
$=-\overrightarrow{\text{AB}}+\overrightarrow{\text{BD}}$
$=-\big(5\vec{\text{a}}+2\vec{\text{b}}\big)+\big(\vec{\text{a}}-3\vec{\text{b}}\big)$
$=-4\vec{\text{a}}-5\vec{\text{b}}$
$\therefore|\overrightarrow{\text{BD}}|=\big|-4\vec{\text{a}}-5\vec{\text{b}}\big|$
$=\big|4\vec{\text{a}}+5\vec{\text{b}}\big|$
$=\sqrt{|4\vec{\text{a}}|^2+|5\vec{\text{b}}|^2+2|4\vec{\text{a}}|\times|5\vec{\text{b}|}\cos\theta}$
$=\sqrt{16|\vec{\text{a}}|^2+25\big|\vec{\text{b}}\big|^2+40\times|\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\frac{\pi}{4}}$
$=\sqrt{16|2\sqrt{2}|^2+25|3|^2+40\times|2\sqrt{2}|\times|3|\times\frac{1}{\sqrt{2}}}$
$=\sqrt{128+25+240}$
$=\sqrt{593}\text{ units}$ 
Therefore, the larger diagonal $=\sqrt{593}$
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Question 841 Mark
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are any three mutualy perpendicular vectors of equal magnitude a, then $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|$ is equal to
  1. $\text{a}$
  2. $\sqrt{2}\text{a}$
  3. $\sqrt{3}\text{a}$
  4. $2\text{a}$
  5. $\text{None of these}$
Answer
  1. $\sqrt{3}\text{a}$
Solution:
Given that
So, $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=|\vec{\text{c}}|=\text{a}\dots(1)$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0\dots(2)$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=\text{a}^2+\text{a}^2+\text{a}^2+0+0+0$ [using (1) and (2)]
$=3\text{a}^2$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}\text{a}$
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Question 851 Mark
If $\vec{\text{r}}.\vec{\text{a}}=\vec{\text{r}}.\vec{\text{b}}=\vec{\text{r}}.\vec{\text{c}}=0$ for some non-zero vector $\vec{\text{r}},$ then the value of $\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big],$ is:
  1. 2
  2. 3
  3. 0
  4. None of these
Answer
  1. 0
Solution:
If $\vec{\text{r}}.\vec{\text{a}}=0$ for some non-zero vector $\vec{\text{r}},$ then either $\vec{\text{a}}$ is a zero - vector or it is perpendicular to $\vec{\text{r}}.$
If one of $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ is zero, then $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
If all $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are non - zero, then they must be coplanar as they are perpendicular to vector $\vec{\text{r}}.$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
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Question 881 Mark
If OACB is a parallelogram with $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$, then $\overrightarrow{\text{OA}}=$
  1. $\big(\vec{\text{a}}+\vec{\text{b}}\big)$
  2. $\big(\vec{\text{a}}-\vec{\text{b}}\big)$
  3. $\frac{1}2\big(\vec{\text{b}}-\vec{\text{a}}\big)$
  4. $\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
Answer
  1. $\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
Solution:
Given a parallelogram OABC such that $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$. Then,
$\overrightarrow{\text{OB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}$
$\Rightarrow\ \overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{BC}}$
$\Rightarrow\ \overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$ $\Big[\because\overrightarrow{\text{BC}}=\overrightarrow{\text{OA}}\Big]$
$\Rightarrow\ \overrightarrow{\text{OB}}=\vec{\text{a}}-\overrightarrow{\text{OA}}\ \dots(1)$
Therefore,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}$
$\Rightarrow\ \overrightarrow{\text{OA}}+\vec{\text{b}}=\vec{\text{a}}-\overrightarrow{\text{OA}}$ [Using (1)]
$\Rightarrow\ 2\overrightarrow{\text{OA}}=\vec{\text{a}}-\vec{\text{b}}$
$\Rightarrow\ \overrightarrow{\text{OA}}=\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
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Question 891 Mark
What is direction of vector $\vec{\text{a}}$ if it is multiplied with $-\lambda$:
  1. Downwards
  2. Upwards
  3. Same
  4. Opposite
Answer
  1. Opposite
Solution:
If the vector is multiplied with $-\lambda$ then its direction become opposite as the direction in which it was previous may be positive or negative. After it is multiplied with a negative value then its direction becomes exactly opposite to the previous direction.
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Question 901 Mark
Choose the correct answer:$\text{Let}\ \vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if,
  1. $\theta=\frac{\pi}{4}$
  2. $\theta=\frac{\pi}{3}$
  3. $\theta=\frac{\pi}{2}$
  4. $\theta=\frac{2\pi}{3}$
Answer
$\text{Let}\ \vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be two unit vectors and $\theta$ be the angle between them.
$\text{Then},\ \big|\vec{\text{a}}\big|=\Big|\vec{\text{b}}\Big|=1.$
$\text{Now},\ \vec{\text{a}}+\vec{\text{b}}$ is a unit vector if $\Big|\vec{\text{a}}+\vec{\text{b}}\Big|=1.$
$\Big|\vec{\text{a}}+\vec{\text{b}}\Big|=1$
$\Rightarrow\Big(\vec{\text{a}}+\vec{\text{b}}\Big)^2=1$
$\Rightarrow\Big(\vec{\text{a}}+\vec{\text{b}}\Big)\cdot\Big(\vec{\text{a}}+\vec{\text{b}}\Big)=1$
$\Rightarrow\vec{\text{a}}.\vec{\text{a}}+\vec{\text{a}}.\vec{\text{b}}+\vec{\text{b}}.\vec{\text{a}}+\vec{\text{b}}.\vec{\text{b}}=1$
$\Rightarrow\Big|\vec{\text{a}}\Big|^2+2\vec{\text{a}}.\vec{\text{b}}+\Big|\vec{\text{b}}\Big|^2=1$
$\Rightarrow1^2+2\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|\cos\theta+1^2=1$
$\Rightarrow1+2.1.1\cos\theta+1=1$
$\Rightarrow\cos\theta=-\frac{1}{2}$
$\Rightarrow\theta=-\frac{2\pi}{3}$
Hence, $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if $\theta=\frac{2\pi}{3}.$
The correct answer is D.
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Question 911 Mark
If $\overline{\text{a}},\overline{\text{b}},\overline{\text{c}}$ are unit vectors such that $\overline{\text{a}}+\overline{\text{b}}+\overline{\text{c}}+\overline{\text{c.a}}=$
  1. $\frac{3}{2}$
  2. $-\frac{3}{2}$
  3. $\frac{1}{2}$
  4. $-\frac{1}{2}$
Answer
  1. $-\frac{3}{2}$
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Question 921 Mark
The value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big),$ is:
  1. 0
  2. -1
  3. 1
  4. 3
Answer
  1. 1
Solution:
$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big)$
$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.(-\hat{\text{j}})+\hat{\text{k}}.\hat{\text{k}}$
$=|\hat{\text{i}}|^2-|\hat{\text{j}}|^2+|\hat{\text{k}}|^2$
$=1-1+1$
$=1$
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Question 931 Mark
Area of a rectangle having vertices A, B, C and D with position vectors $-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k},\ \hat{i}+\frac{1}{2}\hat{j}+4\hat{k},$ $\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}\ \text{and} \ -\hat{i}-\frac{1}{2}\hat{j}+4\hat{k},$
  1. $\frac{1}{2}$
  2. 1
  3. 2
  4. 4
Answer
$\text{Given:}\ \text{ABCD is a rectangle}$ $ \text{Now} \ \ \overrightarrow{\text{AB}}\ $  = Position vector of point B - Position vector of point A$=\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}-\bigg(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\bigg)$ $=\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}+\hat{i}-\frac{1}{2}\hat{j}-4\hat{k}$
$=2\hat{i}+0\hat{j}+0\hat{k}$
$\therefore \ \text{AB}=\bigg|{\overrightarrow{\text{AB}}\bigg|}\ \ =\sqrt{4+0+0}=\sqrt{4}=2$ $\text{And}\ \ \overrightarrow{\text{AD}}$ = Position vector of point D - Position vector of point A $=-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}-\bigg(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\bigg)$ $=-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}+\hat{i}-\frac{1}{2}\hat{j}-4\hat{k}$
$=0\hat{i}-\hat{j}+0\hat{k}$
$\therefore\ \text{AD}=\bigg|\overrightarrow{\text{AD}}\bigg|=\sqrt{0+1+0}=\sqrt{1}=1$ $\therefore\ $ Area of rectangle ABCD = Length x Breadth = AB × AD = 2 × 1 = 2 eq. units Therefore, option (C) is correct.
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Question 941 Mark
If $\vec{\text{a}} $ lies in the plane of vectors $\vec{\text{b}}$ and $\vec{\text{c}},$ then which of the following is correct?
  1. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
  2. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=1$
  3. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=3$
  4. $\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=1$
Answer
  1. ​​​​​​$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
Solution:
If $\vec{\text{a}}$ lies in the plane of vectors $\vec{\text{b}}$ and $\vec{\text{c}},$ then $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ will lie in the same plane, i.e. they will be coplanar.
$\therefore \big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
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Question 951 Mark
Four forces act on a point object. The object will be in equilibrium, if:
  1. All of them are in the same plane
  2. They are opposite to each other in pairs
  3. The sum of x, y and z-components of forces zero separately
  4. They form a closed figure of 4 sides when added as Polygon law
Answer
  1. They form a closed figure of 4 sides when added as Polygon law
Solution:
The equilibrium condition is obtained when the net force acting on the body is zero.
and a closed polygon of 4 sides will give the resultant force as zero and forcing acting will be in the same plane.
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Question 961 Mark
A unit vector along the direction $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ has a magnitude:
  1. $\sqrt{3}$
  2. $\sqrt{2}$
  3. $1$
  4. $0$
Answer
  1. $1$
Solution:
A unit vector along any direction always has magnitude.
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Question 971 Mark
The unit vector perpendicular to the plane passing through points $\text{P}\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big),\text{Q}\big(2\hat{\text{i}}-\hat{\text{k}}\big)$ and $\text{R}\big(2\hat{\text{j}}+\hat{\text{k}}\big)$ is:
  1. $2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
  2. $\sqrt{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
  3. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
  4. $\frac{1}{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
Answer
  1. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
Solution:
The vector $\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$ is perpendicular to the vectors $\overrightarrow{\text{PQ}}$  and $\overrightarrow{\text{PR}}.$
$\therefore$ Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
Now,
$\overrightarrow{\text{PQ}}=\text{P.V}\text{ of }\text{Q}-\text{P.V}.\text{ of P}$
$=\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$\overrightarrow{\text{PR}}=\text{P.V}\text{ of }\text{R}-\text{P.V}.\text{ of P}$
$=-\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
$\therefore\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&-3\\-1&3&-1 \end{vmatrix}$
$=8\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$
$=4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\Rightarrow\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|=\sqrt{64+16+16}$
$=\sqrt{96}$
$=4\sqrt{6}$
Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
$=\frac{4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{4\sqrt{6}}$
$=\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
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Question 981 Mark
If a, b, c are unit vectors such that a + b + c = 0, then the value of a.b + b.c + c.a is:
  1. 2
  2. 3
  3. $-\frac{3}{2}$
  4. None of these
Answer
  1. $-\frac{3}{2}$
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Question 991 Mark
$\mid\text{a}\times\text{b}\mid^2+\mid\text{a.b}\mid^2=144$ and $\mid\text{a}\mid=4$ then $\mid\text{b}\mid$ is equal to:
  1. 12
  2. 3
  3. 8
  4. 4
Answer
  1. 3
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Question 1001 Mark
Choose the correct answer
If $\theta$ is the angle between two vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}},\ \text{then}\ \vec{\text{a}}\cdot\vec{\text{b}}\geq0$ only when,
  1. $0<\theta<\frac{\pi}{2}$
  2. $0\leq\theta\leq\frac{\pi}{2}$
  3. $0<\theta<\pi$
  4. $0\leq\theta\leq\pi$​​
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 $|\vec{\text{a}}|\ \text{and}\ \Big|\vec{\text{b}}\Big|$ are positive
It is known that $\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}| \Big|\vec{\text{b}}\Big|\cos\theta.$
$\therefore\vec{\text{a}}\cdot\vec{\text{b}}\geq0$
$\Rightarrow|\vec{\text{a}}| \Big|\vec{\text{b}}\Big|\cos\theta\geq0$
$\Rightarrow\cos\theta\geq0\ \ \ \Big[\big|\vec{\text{a}}\big|\ \text{and}\ \Big|\vec{\text{b}}\Big|\ \text{are positive}\Big]$
$\Rightarrow0\leq\theta\leq\frac{\pi}{2}$
Hence, $\vec{\text{a}}.\vec{\text{b}}\geq0\ \text{when}\ 0\leq\theta\leq\frac{\pi}{2}.$
The correct answer is B.
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