MCQ 1011 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:
- A
$\frac{3\vec{\text{a}}-2\vec{\text{b}}}{2}$
- B
$\frac{7\vec{\text{a}}-8\vec{\text{b}}}{4}$
- C
$\frac{3\vec{\text{a}}}{4}$
- ✓
$\frac{5\vec{\text{a}}}{4}$
AnswerCorrect option: D. $\frac{5\vec{\text{a}}}{4}$
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 $\text{C}=\frac{3(\vec{\text{a}}+\vec{\text{b}})+1(\vec{2\text{a}}-3\vec{\text{b}})}{3+1}$
$\Rightarrow\text{C}=\frac{5\vec{\text{a}}}{4}$
View full question & answer→MCQ 1021 Mark
Four forces act on a point object. The object will be in equilibrium, if:
- A
All of them are in the same plane
- B
They are opposite to each other in pairs
- C
The sum of x, y and z-components of forces zero separately
- ✓
They form a closed figure of 4 sides when added as Polygon law
AnswerCorrect option: D. They form a closed figure of 4 sides when added as Polygon law
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.
View full question & answer→MCQ 1031 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 $\vec{\alpha}=$
- ✓
$\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
- B
$\frac{2}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
- C
$\frac{1}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
- D
$\frac{1}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
AnswerCorrect option: A. $\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
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
$\text{a}_1=\frac{3}{2};\text{a}_2=\frac{3}{2};\text{a}_3=0$
$\therefore\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$=\frac{3}{2}\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+0\hat{\text{k}}$
$=\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
View full question & answer→MCQ 1041 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:
- A
$\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
- B
$\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{3}$
- ✓
$3(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
- D
$9(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
AnswerCorrect option: C. $3(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
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 $9\cdot\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=9\cdot\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{3}$
$=3(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})$
View full question & answer→MCQ 1051 Mark
The vector $(\cos\text{a}\cos\beta)\hat{\text{i}}+(\cos\text{a}\sin\beta)\hat{\text{j}}+(\sin\text{a})\hat{\text{k}}$is a:
AnswerLet $\vec{\text{a}}=(\cos\text{a}\cos\beta)\hat{\text{i}}+(\cos\text{a}\sin\beta)\hat{\text{j}}+(\sin\text{a})\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{\cos^2\text{a}\cos^2\beta+\cos^2\text{a}\sin^2\beta+\sin^2\text{a}}$
$=\sqrt{\cos^2\text{a}(\cos^2\beta+\sin^2\beta)+\sin^2\text{a}}$
$=\sqrt{\cos^2\text{a}(1)+\sin^2\text{a}}$
$=\sqrt{\cos^2\text{a}+\sin^2\text{a}}$
$=\sqrt{1}$
$=1$
So, $\vec{\text{a}}$ is a unit vector.
View full question & answer→MCQ 1061 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:
- ✓
Exactly two values of (p, q)
- B
More than but not all values of (p, q)
- C
- D
Exactly one values of (p, q)
AnswerCorrect option: A. Exactly two values of (p, q)
Exactly two values of (p, q)
View full question & answer→MCQ 1071 Mark
If $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|,$ then $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=$
AnswerGiven that
$|\vec{\text{a}}|=|\vec{\text{a}}|$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$|\vec{\text{a}}^2-|\vec{\text{a}}|^2$
$=0$
View full question & answer→MCQ 1081 Mark
Which of the following represents coinitial vector:
MCQ 1091 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:
AnswerGiven condition
PR = 5PQ
R − P = 5(Q − P)
R = 5Q − 5P + P
R = 5Q − 4P
R = 5b − 4a
View full question & answer→MCQ 1101 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):
View full question & answer→MCQ 1111 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:
AnswerCorrect option: C. $\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
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 $\frac{\pi}{6}\big)$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\frac{1}{2}\big)^2$
$=\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
View full question & answer→MCQ 1121 Mark
For non-zero vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good, if:
- A
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=0$
- B
$\vec{\text{a}}.\vec{\text{b}}=0=\vec{\text{c}}.\vec{\text{a}}$
- ✓
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
- D
$\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
AnswerCorrect option: C. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
we have
$\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|\big|\cos\theta\big|$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|$ (If $\theta=0^\circ\text{or}180^\circ,$ i.e. vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{c}}$ are parallel)
$=\big|\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\alpha\big)\big|\big|\vec{\text{c}}\big|$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ (If $\alpha=90^\circ,$ i. e. vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular)
$\therefore\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ (If vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other)
Thus, the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good if $\vec{\text{a}}.\vec{\text{b}}=0,\vec{\text{b}}.\vec{\text{c}}=0$ and $\vec{\text{c}}.\vec{\text{a}}=0.$
View full question & answer→MCQ 1131 Mark
Line passing through (3, 4, 5) and (4, 5, 6) has direction ratios:
- ✓
$1,1,1$
- B
$\sqrt{3},\sqrt{3},\sqrt{3}$
- C
$\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
- D
$7,9,11$
AnswerCorrect option: A. $1,1,1$
Given points (3, 4, 5) and (4, 5, 6) The drs are given as (4 - 3, 5 - 4, 6 - 5) = (1, 1, 1)
View full question & answer→MCQ 1141 Mark
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$ is equal to:
AnswerWe have
$\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
$\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}+\frac{\big[\vec{\text{b}}\vec{\text{a}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}$ (By definition of scalar tiple product)
$=\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}+\frac{-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}$ (Change in cyclic order of vectors changes the sign of the scalar triple product)
$=1-1$
$=0$
View full question & answer→MCQ 1151 Mark
Answer
Zero vector, is a vector of length 0, and thus has all components equal to zero. It is the additive identity of the additive group of vectors.
Thus, it has zero magnitude and arbitrary direction.
View full question & answer→MCQ 1161 Mark
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)=$
- A
$0$
- B
$-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
- C
$2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
- ✓
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
AnswerCorrect option: D. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
We have
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}.\big[\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{a}}+\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{b}}+\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{c}}\big]$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}+0+\vec{\text{c}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+0$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}-\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)+\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$
$=0+0+0+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
View full question & answer→MCQ 1171 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:
AnswerCorrect option: A. $\pm 1$
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 $-\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ $\big(\therefore\big|\vec{\text{c}}\big|=1,\cos0^\circ=1\text{ and }\cos180^\circ=-1\big)$ $$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin90^\circ$ or $-\big|\text{a}\big|\big|\vec{\text{b}}\big|\sin90$ $\big(\therefore\vec{\text{a}} \text{ is perpendicular to }\vec{\text{b}})$
$=1 \text{ or }-1$ $\big(\therefore\big|\vec{\text{a}}\big|=1 \text{ and }\big|\vec{\text{b}}\big|=1\big)$ $$
$=\pm1$
View full question & answer→MCQ 1181 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:
- A
$\text{m}=\text{x}$
- B
$\text{x}=\mid{\text{m}}\mid$
- ✓
$\text{x}=\frac{1}{\mid\text{m}\mid}$
- D
$\text{x}=\frac{\text{m}}{2}$
AnswerCorrect option: C. $\text{x}=\frac{1}{\mid\text{m}\mid}$
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.
View full question & answer→MCQ 1191 Mark
Point (4, 0) lies on ________?
- A
$\vec{\text{XO}}$
- B
$\vec{\text{YO}}$
- ✓
$\vec{\text{OX}}$
- D
$\vec{\text{OY}}$
AnswerCorrect option: C. $\vec{\text{OX}}$
$\vec{\text{OX}}$
View full question & answer→MCQ 1201 Mark
Let the vectors $\vec{a}\ \text{and}\ \vec{b}$ be such that $|\vec{a}|=3\ \text{and}\ \big|\vec{b}\big|=\frac{\sqrt{2}}{3},\ \text{then}\ \vec{a}\times\vec{b}$ is a unit vector, if the angle between $\vec{a}\ \text{and}\ \vec{b}\ \text{is}$
- A
$\pi/6$
- ✓
$\pi/4$
- C
$\pi/3$
- D
$\pi/2$
AnswerCorrect option: B. $\pi/4$
$\text{Given:}\ \ \big|\vec{a}\big|=3,\big|\vec{b}\big|=\frac{\sqrt{2}}{3}\ \text{and}\ \vec{a}\times\vec{b}$ is a unit vector.
$\Rightarrow\ \ \big|\vec{a}\times\vec{b}\big|=1\ $ $\Rightarrow\ \big|\vec{a}\big|.\Big|\vec{b}\Big|\ \text{sin}\ \theta=1,\ \text{where}\ \theta\ \text{is the angle between}\ \vec{a}\ \text{and}\ \vec{b}.$
$\Rightarrow\ \ 3\bigg(\frac{\sqrt{2}}{3}\bigg)\ \text{sin}\ \theta=1$ $\Rightarrow\ \sqrt{2}\ \text{sin}\ \theta=1\ \Rightarrow\ \ \text{sin}\ \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow\ \text{sin}\ \theta=\text{sin}\frac{\pi}{{4}}\ \ \Rightarrow\ \theta=\frac{\pi}{4}$
Therefore, option (B) is correct.
View full question & answer→MCQ 1211 Mark
Choose the correct answer from the given four options.
The angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ is:
- A
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{3}$
- C
$\frac{\pi}{2}$
- D
$\frac{5\pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{3}$
Here, $|\vec{\text{a}}|=\sqrt{3},|\vec{\text{b}}|=4$ and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ [given]
We know that, $\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\Rightarrow2\sqrt{3}=\sqrt{3}.4.\cos\theta$
$\Rightarrow\cos\theta=\frac{2\sqrt{3}}{4\sqrt{3}}=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}$
View full question & answer→MCQ 1221 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,
- ✓
$(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
- B
$(1+\sqrt3)\hat{\text{i}}+(1-\sqrt3)\hat{\text{j}}$
- C
$(1+\sqrt3)\hat{\text{i}}+(\sqrt3-1)\hat{\text{j}}$
- D
AnswerCorrect option: A. $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
$\Big|\overrightarrow{\text{AO}}\Big|=2\sqrt2$
$\Big|\overrightarrow{\text{AO}}\Big|=\Big|\overrightarrow{\text{BO}}\Big|=\Big|\overrightarrow{\text{CO}}\Big|=2\sqrt2=\text{R}$
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 $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
$\overrightarrow{\text{AO}}=(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$ View full question & answer→MCQ 1231 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 =
AnswerLet
$\vec{\text{a}}=4\hat{\text{i}}+11\hat{\text{j}}+{\text{m}}\hat{\text{k}}$
$\vec{\text{b}}=7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
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. $\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}4 11 \text{m}7 2 61 5 4 \end{vmatrix}=0$
$\Rightarrow 4(8-30)-11(28-6)+\text{m}(35-2)=0$
$\Rightarrow-88-242+33\text{m}=0$
$\Rightarrow33\text{m}=330$
$\therefore\text{m}=10$
View full question & answer→MCQ 1241 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:
AnswerIf $\vec{\text{r}}.\vec{\text{a}}=0$ 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 $\vec{\text{r}}.$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
View full question & answer→MCQ 1251 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:
AnswerCorrect option: D. $\frac{2\pi}{3}<\theta<\pi$
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 $\pi.$
View full question & answer→MCQ 1261 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?
MCQ 1271 Mark
The position vector of the point (1, 2, 0) is:
MCQ 1281 Mark
Two vectors each of magnitudes $1$ unit are inclined at $60^{\circ}$ to each other. The difference of the vectors has a magnitude $...........?$
- A
$0$ units
- B
$1$ units
- ✓
$2$ units
- D
$3$ units
AnswerCorrect option: C. $2$ units
View full question & answer→MCQ 1291 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:
AnswerLet $\theta$ 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,
$\cos\theta>0$
$\Rightarrow\text{x}^2-3\text{x}-28>0$
$\Rightarrow(\text{x}-7)(\text{x}+4)>0$
$\Rightarrow\text{x}\in(-\infty,-4)\cup(7,\infty)$
$\Rightarrow\text{x}\in\text{R}-[-4,7]$
View full question & answer→MCQ 1301 Mark
If the position vectors of P and Q are $\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}$ and $5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$ then the cosine of the angle getween $\overrightarrow{\text{PQ}}$ and y-axis is:
- A
$\frac{5}{\sqrt{162}}$
- B
$\frac{4}{\sqrt{162}}$
- ✓
$-\frac{5}{\sqrt{162}}$
- D
$\frac{11}{\sqrt{162}}$
AnswerCorrect option: C. $-\frac{5}{\sqrt{162}}$
$\overrightarrow{\text{PQ}}=\overrightarrow{\text{OQ}}-\overrightarrow{\text{OP}}=5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}-\big(\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}\big)=4\hat{\text{i}}-5\hat{\text{j}}+11\hat{\text{k}}$
The unit vector along y-axis is $\hat{\text{j}}.$
Let $\theta$ be the required angle.
$\cos\theta=\frac{\overrightarrow{\text{PQ}}.\hat{\text{j}}}{\big|\overrightarrow{\text{PQ}}\big|\big|\hat{\text{j}}\big|}=\frac{-5}{\sqrt{16+25+121}\sqrt{1}}=\frac{-5}{\sqrt{162}}$
View full question & answer→MCQ 1311 Mark
find the coordinate of the tip of the position vector which is equivalent to $\overrightarrow{\text{AB}}$ where the coordinates of A and B are (-1, 3) and (-2, 1) respectively:
View full question & answer→MCQ 1321 Mark
If a + b + c = 0, then a × b =
MCQ 1331 Mark
The orthogonal projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is:
- A
$\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{a}}}{|\vec{\text{a}}|^2}$
- ✓
$\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
- C
$\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
- D
$\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
AnswerCorrect option: B. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
The orthogonal projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$
View full question & answer→MCQ 1341 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:
AnswerCorrect option: C. $-\frac{3}{2}.$
We have $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})=0$
$\Rightarrow\vec{\text{a}}^2+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}^2+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{b}}+\vec{\text{c}}^2=0$
$\Rightarrow\vec{\text{a}}^2+\vec{\text{b}}^2+\vec{\text{c}}^2+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$ $[\because\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}},\vec{\text{b}}\cdot\vec{\text{c}}=\vec{\text{c}}\cdot\vec{\text{b}}\text{ and }\vec{\text{c}}\cdot\vec{\text{a}}=\vec{\text{a}}\cdot\vec{\text{c}}]$
$\Rightarrow1+1+1+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$
$\Rightarrow\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}=-\frac{3}{2}$
View full question & answer→MCQ 1351 Mark
Which of the following represents equal vectors:
MCQ 1361 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:
- A
$10\hat{\text{i}}+2\hat{\text{j}}+11\hat{\text{k}}$
- ✓
$10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
- C
$10\hat{\text{i}}-3\hat{\text{j}}+11\hat{\text{k}}$
- D
$10\hat{\text{i}}-2\hat{\text{j}}-10\hat{\text{k}}$
AnswerCorrect option: B. $10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&-1\\1&4&-2 \end{vmatrix}$
$=10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
View full question & answer→MCQ 1371 Mark
A unit vector along the direction $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ has a magnitude:
- A
$\sqrt{3}$
- B
$\sqrt{2}$
- ✓
$1$
- D
$0$
AnswerA unit vector along any direction always has magnitude.
View full question & answer→MCQ 1381 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?
AnswerSo, shortest distance = 24
View full question & answer→MCQ 1391 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.
- A
$\frac{\text{d}}{2\text{v}}$
- ✓
$\frac{\text{d}}{\text{v}}$
- C
$\frac{\text{3d}}{2\text{v}}$
- D
AnswerCorrect option: B. $\frac{\text{d}}{\text{v}}$
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
$\text{t}=\frac{\text{d}}{\text{v}}$
$=\frac{\frac{\text{d}}{\sqrt{2}}}{\frac{\text{v}}{\sqrt{2}}}=\frac{\text{d}}{\text{v}}$
View full question & answer→MCQ 1401 Mark
If $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$then $\vec{\text{a}}=$
AnswerCorrect option: B. $\hat{\text{i}}$
Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
and $\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$
and $\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$
Given,
$\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1$
$\Rightarrow\text{a}_1=\text{a}_1+\text{a}_2=\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1,\text{a}_2=0,\text{a}_3=0$
So, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$
View full question & answer→MCQ 1411 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:
- ✓
$2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
- B
$\sqrt{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
- C
$\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
- D
$\frac{1}{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
AnswerCorrect option: A. $2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
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 $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
$=\frac{4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{4\sqrt{6}}$
$=\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→MCQ 1421 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:
AnswerCorrect option: C. $-\frac{3}{2}$
$-\frac{3}{2}$
View full question & answer→MCQ 1431 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 $\mid\vec{\text{x}}+2\vec{\text{y}}\mid=$
- A
$40$
- B
$\sqrt{35}$
- C
$\sqrt{17}$
- ✓
$2\sqrt{10}$
AnswerCorrect option: D. $2\sqrt{10}$
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 $\mid\vec{\text{x}}+2\vec{\text{y}}\mid=\mid6\hat{\text{i}}-2\hat{\text{j}}\mid=\sqrt{40}=2\sqrt{10}$
View full question & answer→MCQ 1441 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:
- A
$\cos \theta = \frac{4}{5}$
- B
$\sin \theta = \frac{1}{\sqrt{2}}$
- ✓
$\cos \theta = -\frac{4}{5}$
- D
$\cos \theta = -\frac{3}{5}$
AnswerCorrect option: C. $\cos \theta = -\frac{4}{5}$
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
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\frac{-8}{9}=\frac{\sqrt{10}}{3}\times\frac{\sqrt{10}}{3}\cos\theta$
$\Rightarrow\frac{-8}{9}=\frac{10}{9}\cos\theta$
$\Rightarrow\cos\theta=\frac{-8}{9}\times\frac{9}{10}=\frac{-4}{5}$
View full question & answer→MCQ 1451 Mark
The magnitude of the vector 6i + 2j + 3k is equal to:
MCQ 1461 Mark
If OACB is a parallelogram with $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$, then $\overrightarrow{\text{OA}}=$
- A
$\big(\vec{\text{a}}+\vec{\text{b}}\big)$
- B
$\big(\vec{\text{a}}-\vec{\text{b}}\big)$
- C
$\frac{1}2\big(\vec{\text{b}}-\vec{\text{a}}\big)$
- ✓
$\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
AnswerCorrect option: D. $\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
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)]
$\Rightarrow\ 2\overrightarrow{\text{OA}}=\vec{\text{a}}-\vec{\text{b}}$
$\Rightarrow\ \overrightarrow{\text{OA}}=\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
View full question & answer→MCQ 1471 Mark
The scalar product of 5i + j - 3k and 3i - 4j + 7k is:
AnswerLet 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
View full question & answer→MCQ 1481 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:
AnswerHere, $|\vec{{\text{a}}}|=10,|\vec{{\text{b}}}|=2$ and $\vec{{\text{a}}}\cdot\vec{\text{b}}=12$ [given]
$\therefore\vec{{\text{a}}}\cdot\vec{\text{b}}=|\vec{{\text{a}}}||\vec{{\text{b}}}|\cos\theta$
$12=10\times2\cos\theta$
$\Rightarrow\cos\theta=\frac{12}{20}=\frac{3}{5}$
$\Rightarrow\sin\theta=\sqrt{1-\cos\theta}$
$=\sqrt{1-\frac{9}{25}}$
$\sin\theta=\pm\frac{4}{5}$
$\therefore|\vec{{\text{a}}}\times\vec{{\text{b}}}|=|\vec{{\text{a}}}||\vec{{\text{b}}}||\sin\theta|$
$=10\times2\times\frac{4}{5}$
$=16$
View full question & answer→MCQ 1491 Mark
In triangle ABC (Fig 10.18), which of the following is not true:
- A
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec{0}$
- B
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec{0}$
- ✓
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec{0}$
- D
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec{0}$
AnswerCorrect option: C. $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec{0}$
On applying the triangle law of addition in the given triangle, we have:
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\ \ \ \ \ \ \ \ \ ....(1)$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=-\overrightarrow{\text{CA}}$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec{0}\ \ \ \ ....(2)$
$\therefore$ The equation given in alternative A is true.
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec{0}$
$\therefore$ The equation given in alternative B is true.
From equation (2), we have:
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec{0}$
$\therefore$ The equation given in alternative D is true.
Now, consider the equation given in alternative C:
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec{0}$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{CA}}\ \ \ \ ....(3)$
From equations (1) and (3), we have:
$\overrightarrow{\text{AC}}=\overrightarrow{\text{CA}}$
$\Rightarrow\overrightarrow{\text{AC}}=-\overrightarrow{\text{AC}}$
$\Rightarrow\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}=\vec{0}$
$\Rightarrow2\overrightarrow{\text{AC}}=\vec{0}$
$\Rightarrow\overrightarrow{\text{AC}}=\vec{0},$ which is not true.
Hence, the equation given in alternative C is incorrect.
The correct answer is C. View full question & answer→MCQ 1501 Mark
If $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$ and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}.\vec{\text{a}}\neq0,$ then:
AnswerCorrect option: A. $\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}-\vec{\text{a}}.\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}.\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\big(\vec{\text{b}}-\vec{\text{c}}\big)$
$|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta\dots(1)$
and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Then, $|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0\dots(2)$
Here, it is given that $\vec{\text{a}}\neq0$
Therefore, for eq. (1) and eq. (2) to be 0
We have,
$\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0$
For $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0,$ one of $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|$ or $\cos\theta$ must be 0
Case 1 :
Let $\cos\theta=0$
$\Rightarrow\theta=90^\circ$
$\Rightarrow\sin\theta=1$
& if $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0$ and $\sin\theta=1$
Then $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Case 2 :
Let $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Hence, $\vec{\text{b}}=\vec{\text{c}}$
View full question & answer→