MCQ 2512 Marks
The direction cosines of the vector $2 \hat{i}+2 \hat{j}-\hat{k}$ are
- A
$\frac{2}{3}, \frac{1}{3}, \frac{1}{3}$
- B
$\frac{1}{3}, \frac{1}{3}, \frac{2}{3}$
- C
$\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}$
- ✓
$\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}$
AnswerCorrect option: D. $\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}$
(D) Let $\overline{ r }=2 \hat{ i }+2 \hat{ j }-\hat{ k }$
$|\overline{ r }|=\sqrt{2^2+2^2+(-1)^2}=3$
$\therefore \quad$ The d.c.s are $\frac{x}{|\overline{ r }|}, \frac{y}{|\overline{ r }|}, \frac{ z }{|\overline{ r }|}$
i.e., $\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}$
View full question & answer→MCQ 2522 Marks
The co-ordinates of the point P are $(x, y, z)$ and the direction cosines of the line OP when O is the origin, are $l, m, n$. If $OP = r$, then
AnswerCorrect option: C. $x=l r , y= mr , z = nr$
View full question & answer→MCQ 2532 Marks
A line makes angles of $45^{\circ}$ and $60^{\circ}$ with the positive axes of X and Y respectively. The angle made by the same line with the positive axis of $Z$ is
- A
$30^{\circ}$ or $60^{\circ}$
- B
$60^{\circ}$ or $90^{\circ}$
- C
$90^{\circ}$ or $120^{\circ}$
- ✓
$60^{\circ}$ or $120^{\circ}$
AnswerCorrect option: D. $60^{\circ}$ or $120^{\circ}$
(D) Since $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$\therefore \cos ^2 45^{\circ}+\cos ^2 60^{\circ}+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \gamma=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
$\Rightarrow \cos \gamma= \pm \frac{1}{2}$
$\Rightarrow \gamma=60^{\circ}$ or $120^{\circ}$
View full question & answer→MCQ 2542 Marks
If a line makes angles $45^{\circ}, 60^{\circ}$ and $60^{\circ}$ with the $X , Y$ and Z -axes respectively, then its direction cosines are
- ✓
$\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$
- B
$0, \frac{1}{2}, \frac{1}{2}$
- C
$\frac{1}{\sqrt{2}}, 0, \frac{1}{2}$
- D
$\frac{1}{2}, 0, \frac{1}{\sqrt{2}}$
AnswerCorrect option: A. $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$
(A) Let the direction cosines of the line be $l, m, n$
$\therefore \quad l=\cos 45^{\circ}, m =\cos 60^{\circ}, n =\cos 60^{\circ}$
$\Rightarrow l=\frac{1}{\sqrt{2}}, m=\frac{1}{2}$ and $n =\frac{1}{2}$
∴ d.c.s are $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$.
View full question & answer→MCQ 2552 Marks
Which of the following angles made by a line with co-ordinate axes are not possible?
- A
$30^{\circ}, 60^{\circ}, 90^{\circ}$
- ✓
$30^{\circ}, 45^{\circ}, 60^{\circ}$
- C
$45^{\circ}, 90^{\circ}, 45^{\circ}$
- D
$60^{\circ}, 45^{\circ}, 60^{\circ}$
AnswerCorrect option: B. $30^{\circ}, 45^{\circ}, 60^{\circ}$
(B) For option (B),
$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma \neq 1$
∴ option (B) is correct answer.
View full question & answer→MCQ 2562 Marks
If the direction cosines of a line are $\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$, then
- A
$c>0$
- ✓
$c= \pm \sqrt{3}$
- C
$0< c <1$
- D
$c>2$
AnswerCorrect option: B. $c= \pm \sqrt{3}$
(B) $l^2+ m ^2+ n ^2=1$
$\therefore \quad \frac{1}{ c ^2}+\frac{1}{ c ^2}+\frac{1}{ c ^2}=1$
$\Rightarrow c^2=3 \Rightarrow c= \pm \sqrt{3}$
View full question & answer→MCQ 2572 Marks
If $\left(\frac{1}{2}, \frac{1}{3}, n \right)$ are the direction cosines of a line, then the value of $n$ is
- ✓
$\frac{\sqrt{23}}{6}$
- B
$\frac{23}{6}$
- C
$\frac{2}{3}$
- D
$\frac{3}{2}$
AnswerCorrect option: A. $\frac{\sqrt{23}}{6}$
(A) $l^2+ m ^2+ n ^2=1$
$\therefore\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+ n ^2=1$
$\Rightarrow n ^2=\frac{23}{36} \Rightarrow n = \pm \frac{\sqrt{23}}{6}$
View full question & answer→MCQ 2582 Marks
If $\alpha, \beta, \gamma$ be the angles which a line makes with the co-ordinate axes, then
- A
$\sin ^2 \alpha+\cos ^2 \beta+\sin ^2 \gamma=1$
- ✓
$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
- C
$\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=1$
- D
$\cos ^2 \alpha+\cos ^2 \beta+\sin ^2 \gamma=1$
AnswerCorrect option: B. $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
View full question & answer→MCQ 2592 Marks
The direction cosines of Y -axis are
- ✓
$0,1,0$
- B
$1,0,0$
- C
$0,0,1$
- D
$0,1,1$
AnswerCorrect option: A. $0,1,0$
(A) The d.c.s of Y -axis are $\cos 90^{\circ}, \cos 0^{\circ}, \cos 90^{\circ}$
i.e. $0,1,0$
View full question & answer→MCQ 2602 Marks
The direction cosines of X -axis are
- ✓
$1,0,0$
- B
$0,1,0$
- C
$0,0,1$
- D
$1,1,1$
AnswerCorrect option: A. $1,0,0$
(A) The d.c.s of X -axis are $1,0,0$.
View full question & answer→MCQ 2612 Marks
If $\overline{ a }$ makes an acute angle with $\overline{ b }, \overline{ r } \cdot \overline{ a }=0$ and $\overline{ r } \times \overline{ b }=\overline{ c } \times \overline{ b }$, then $\overline{ r }=$
- A
$\bar{a} \times \bar{c}-\bar{b}$
- B
$\bar{c} \times \bar{a}$
- ✓
$\bar{c}-\left(\frac{\bar{c} \cdot \bar{a}}{\bar{b} \cdot \bar{a}}\right) \bar{b}$
- D
$\bar{c}+\left(\frac{\bar{c} \cdot \bar{a}}{\bar{b} \cdot \bar{a}}\right) \bar{b}$
AnswerCorrect option: C. $\bar{c}-\left(\frac{\bar{c} \cdot \bar{a}}{\bar{b} \cdot \bar{a}}\right) \bar{b}$
(C) $\overline{ r } \times \overline{ b }=\overline{ c } \times \overline{ b }$
$\begin{array}{l}\Rightarrow \overline{ a } \times(\overline{ r } \times \overline{ b })=\overline{ a } \times(\overline{ c } \times \overline{ b }) \\ \Rightarrow(\overline{ a } \cdot \overline{ b }) \overline{ r }-(\overline{ a } \cdot \overline{ r }) \overline{ b }=(\overline{ a } \cdot \overline{ b }) \overline{ c }-(\overline{ a } \cdot \overline{ c }) \overline{ b } \\ \Rightarrow(\overline{ a } \cdot \overline{ b }) \overline{ r }-0=(\overline{ a } \cdot \overline{ b }) \overline{ c }-(\overline{ a } \cdot \overline{ c }) \overline{ b } \\ \Rightarrow \overline{ r }=\overline{ c }-\left(\frac{\overline{ c } \cdot \overline{ a }}{\overline{ b } \cdot \overline{ a }}\right) \overline{ b }\end{array}$
View full question & answer→MCQ 2622 Marks
Let $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$. Then, the vector $\overline{ b }$ satisfying $\overline{ a } \times \overline{ b }+\overline{ c }=0$ and $\overline{ a } \cdot \overline{ b }=3$, is
- ✓
$-\hat{i}+\hat{j}-2 \hat{k}$
- B
$2 \hat{ i }-\hat{ j }+2 \hat{ k }$
- C
$\hat{ i }-\hat{ j }-2 \hat{ k }$
- D
$\hat{ i }+\hat{ j }-2 \hat{ k }$
AnswerCorrect option: A. $-\hat{i}+\hat{j}-2 \hat{k}$
(A) Given, $\bar{a} \times \bar{b}+\bar{c}=0$
$\Rightarrow \overline{ a } \times(\overline{ a } \times \overline{ b })+\overline{ a } \times \overline{ c }=0$
$\begin{array}{l}\Rightarrow(\bar{a} \cdot \bar{b}) \bar{a}-(\bar{a} \cdot \bar{a}) \bar{b}+\bar{a} \times \bar{c}=0 \\ \Rightarrow 3 \bar{a}-2 \bar{b}+\bar{a} \times \bar{c}=0 \Rightarrow 2 \bar{b}=3 \bar{a}+\bar{a} \times \bar{c} \\ \Rightarrow 2 \bar{b}=3 \hat{j}-3 \hat{k}-2 \hat{i}-\hat{j}-\hat{k}=-2 \hat{i}+2 \hat{j}-4 \hat{k} \\ \Rightarrow \bar{b}=-\hat{i}+\hat{j}-2 \hat{k}\end{array}$
View full question & answer→MCQ 2632 Marks
If the volume of tetrahedron where vertices $(1,2,0)$,$(2,0,4),(-1,2,0)$ and $(-1,1, \lambda)$ is $\frac{2}{3} cu$ unit, then the value of $\lambda$ is
Answer(B) Let $A \equiv(1,2,0), B \equiv(2,0,4), C \equiv(-1,2,0)$ and $D \equiv(-1,1, \lambda)$ be the vertices of the tetrahedron
$\therefore \quad \overline{ AB }=\hat{ i }-2 \hat{ j }+4 \hat{ k }$
$\overline{ AC }=-2 \hat{ i }$
$\overline{ AD }=-2 \hat{ i }-\hat{ j }+\lambda \hat{ k }$
Volume of tetrahedron $=\frac{1}{6}$ $[\overline{ AB }$ $\overline{ AC }$ $\overline{ AD }]$
$\Rightarrow \frac{2}{3}=\frac{1}{6}\left|\begin{array}{ccc}1 & -2 & 4 \\ -2 & 0 & 0 \\ -2 & -1 & \lambda\end{array}\right|$
$\Rightarrow 2(-2 \lambda)+4(2)=4$
$\Rightarrow \lambda=1$
View full question & answer→MCQ 2642 Marks
The volume of a tetrahedron (in cubic units) whose vertices are $4 \hat{i}+5 \hat{j}+\hat{k}, -\hat{j}+\hat{k}$, $3 \hat{i}+9 \hat{j}+4 \hat{k}$ and $-2 \hat{i}+4 \hat{j}+4 \hat{k}$ is
Answer(B) Let $A , B , C$ and D be the given points.
$\begin{array}{ll}\therefore \quad & \overline{ AB }=-4 \hat{ i }-6 \hat{ j }, \overline{ AC }=-\hat{ i }+4 \hat{ j }+3 \hat{ k }, \text { and } \\ & \overline{ AD }=-6 \hat{ i }-\hat{ j }+3 \hat{ k }\end{array}$
Volume of tetrahedron $=\frac{1}{6}\left|\begin{array}{ccc}-4 & -6 & 0 \\ -1 & 4 & 3 \\ -6 & -1 & 3\end{array}\right|$
$\begin{array}{l}=\frac{30}{6} \\ =5 cu . \text { units }\end{array}$
View full question & answer→MCQ 2652 Marks
The volume of the parallelopiped whose edges are represented by $-12 \hat{i}+\alpha \hat{k}, 3 \hat{j}-\hat{k}$ and $2 \hat{ i }+\hat{ j }-15 \hat{ k }$ is 546 . Then $\alpha=$
Answer(C) Volume of parallelopiped $=\left|\begin{array}{ccc}-12 & 0 & \alpha \\ 0 & 3 & -1 \\ 2 & 1 & -15\end{array}\right|$
$\begin{array}{l}\Rightarrow 546=-12(-45+1)+\alpha(0-6) \\ \Rightarrow \alpha=-3\end{array}$
View full question & answer→MCQ 2662 Marks
The volume of the parallelopiped whose edges are represented by $\bar{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+2 \hat{k}$, $\overline{ c }=2 \hat{ i }+\hat{ j }-\hat{ k }$ is
Answer(A) Volume of parallelopiped $=\left|\begin{array}{ccc}2 & -3 & 1 \\ 1 & -1 & 2 \\ 2 & 1 & -1\end{array}\right|$
$=2(1-2)+3(-1-4)+1(1+2)=-14$
But, volume cannot be negative.
$\therefore \quad$ Volume of parallelopiped $=14 cu$. units
View full question & answer→MCQ 2672 Marks
If the vectors $2 \hat{i}-3 \hat{j}, \hat{i}+\hat{j}-\hat{k}$ and $3 \hat{i}-\hat{k}$ form three concurrent edges of a parallelopiped, then the volume of the parallelopiped is
Answer(C) Let $\bar{a}=2 \hat{i}-3 \hat{j}, \bar{b}=\hat{i}+\hat{j}-\hat{k}$ and $\bar{c}=3 \hat{i}-\hat{k}$
Volume of parallelopiped $=\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
$=\left|\begin{array}{ccc}2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1\end{array}\right|$
$\begin{array}{l}=2(-1)+3(-1+3) \\ =4 \text { cu.units }\end{array}$
View full question & answer→MCQ 2682 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are non- coplanar vectors, then $\left[\begin{array}{lll}\overline{ a }+2 \overline{b} & \overline{ a }+\overline{ c } & \overline{ b }\end{array}\right]=$
- A
$0$
- B
$[\overline{ a } \overline{ b } \overline{ c }]$
- ✓
$-[\overline{ a } \overline{ b } \overline{ c }]$
- D
$2[\overline{ a } \overline{ b } \overline{ c }]$
AnswerCorrect option: C. $-[\overline{ a } \overline{ b } \overline{ c }]$
(C) $\left[\begin{array}{lll}\overline{ a }+2 \overline{b} & \overline{ a }+\overline{ c } & \overline{ b }\end{array}\right]$
$=\left[\begin{array}{lll}\overline{ a } & \overline{ a }+\overline{ c } & \overline{ b }\end{array}\right]+\left[\begin{array}{lll}2 \overline{b} & \overline{ a }+\overline{ c } & \overline{ b }\end{array}\right]$
$=[\overline{ a } \overline{ a } \overline{ b }]+[\overline{ a } \overline{ c } \overline{ b }]+[2 \overline{b} \overline{ a } \overline{ b }]+[2 \overline{b} \overline{ c } \overline{ b }]$
$\begin{array}{l}=0-[\overline{ a } \overline{ b } \overline{ c }]+2(0)+2(0) \\ =-[\overline{ a } \overline{ b } \overline{ c }]\end{array}$
View full question & answer→MCQ 2692 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are any three coplanar unit vectors, then
- A
$\overline{ a } \cdot(\overline{ b } \times \overline{ c })=1$
- B
$\bar{a} \cdot(\bar{b} \times \bar{c})=3$
- ✓
$(\bar{a} \times \bar{b}) \cdot \bar{c}=0$
- D
$(\bar{c} \times \bar{a}) \cdot \bar{b}=1$
AnswerCorrect option: C. $(\bar{a} \times \bar{b}) \cdot \bar{c}=0$
(C) $\overline{ a } \cdot(\overline{ b } \times \overline{ c })=0$ or $(\overline{ a } \times \overline{ b }) \cdot \overline{ c }=0$
View full question & answer→MCQ 2702 Marks
If $\bar{a}$ and $\bar{b}$ be parallel vectors, then $[\bar{a} \bar{c} \bar{b}]=$
Answer(A) $\left[\begin{array}{lll}\overline{ a } & \overline{ c } & \overline{ b }\end{array}\right]=\overline{ a } \cdot(\overline{ c } \times \overline{ b })$
$=\bar{c} \cdot(\bar{b} \times \bar{a})$
$=0 \quad \ldots .[\because \overline{ a }$ and $\overline{ b }$ are parallel $]$
View full question & answer→MCQ 2712 Marks
$[\bar{a} \bar{b} \bar{a} \times \bar{b}]$ is equal to
AnswerCorrect option: B. $|\overline{ a } \times \overline{ b }|^2$
(B) $\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ a } \times \overline{ b }\end{array}\right]=\overline{ a } \cdot[\overline{ b } \times(\overline{ a } \times \overline{ b })]$
$=(\overline{ a } \times \overline{ b }) \cdot(\overline{ a } \times \overline{ b })$
$=|\overline{ a } \times \overline{ b }|^2$
View full question & answer→MCQ 2722 Marks
$\bar{a} \cdot(\bar{a} \times \bar{b})=$
Answer(C) $\overline{ a } \cdot(\overline{ a } \times \overline{ b })=(\overline{ a } \times \overline{ a }) \cdot b =0$
View full question & answer→MCQ 2732 Marks
If the points having the position vectors $3 \hat{i}-2 \hat{j}-\hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, $-\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are coplanar then $\lambda=$
- A
$-8$
- B
$8$
- C
$\frac{146}{17}$
- ✓
$\frac{-146}{17}$
AnswerCorrect option: D. $\frac{-146}{17}$
(D) Let $\bar{a}=3 \hat{i}-2 \hat{j}-\hat{k}, \bar{b}=2 \hat{i}+3 \hat{j}-4 \hat{k}$, $\bar{c}=-\hat{i}+\hat{j}+2 \hat{k}$ and $\bar{d}=4 \hat{i}+5 \hat{j}+\lambda \hat{k}$
Since the given points are coplanar,
$\left[\begin{array}{lll}\overline{ AB } & \overline{ AC } & \overline{ AD }\end{array}\right]=0$
$\Rightarrow\left|\begin{array}{ccc}-1 & 5 & -3 \\ -4 & 3 & 3 \\ 1 & 7 & \lambda+1\end{array}\right|=0$
$\Rightarrow-1(3 \lambda+3-21)-5(-4 \lambda-4-3)$$-3(-28-3)=0$
$\begin{array}{l}\Rightarrow-3 \lambda+18+20 \lambda+35+93=0 \\ \Rightarrow 17 \lambda=-146 \\ \Rightarrow \lambda=\frac{-146}{17}\end{array}$
View full question & answer→MCQ 2742 Marks
If the vectors $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\overline{ c }=x \hat{ i }+(x-2) \hat{ j }-\hat{ k }$ are coplanar, then $x=$
Answer(D) Since $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\overline{ c }=x \hat{ i }+(x-2) \hat{ j }-\hat{ k }$ are coplanar vectors,
$\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]=0$
$\Rightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1\end{array}\right|=0$
$\begin{array}{l}\Rightarrow 1[1-2(x-2)]-1(-1-2 x)+1(x-2+x)=0 \\ \Rightarrow 1-2 x+4+1+2 x+2 x-2=0 \\ \Rightarrow 2 x=-4 \\ \Rightarrow x=-2\end{array}$
View full question & answer→MCQ 2752 Marks
If the vectors $4 \hat{i}+11 \hat{j}+m \hat{k}, 7 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\hat{ i }+5 \hat{ j }+4 \hat{ k }$ are coplanar, then m is equal to
Answer(C) Let $\bar{a}=4 \hat{i}+11 \hat{j}+m \hat{k}, \quad \bar{b}=7 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\overline{ c }=\hat{ i }+5 \hat{ j }+4 \hat{ k }$
Since $\overline{ a }, \overline{ b }$ and $\overline{ c }$ are coplanar, $[\overline{ a } \overline{ b } \overline{ c }]=0$
.$\Rightarrow\left|\begin{array}{ccc}4 & 11 & m \\ 7 & 2 & 6 \\ 1 & 5 & 4\end{array}\right|=0$
$\begin{array}{l}\Rightarrow 4(8-30)-11(28-6)+ m (35-2)=0 \\ \Rightarrow-330+33 m=0 \\ \Rightarrow m=10\end{array}$
View full question & answer→MCQ 2762 Marks
If a vector $\bar{\alpha}$ lie in the plane $\bar{\beta}$ and $\bar{\gamma}$, then which is correct?
- ✓
$\left[\begin{array}{lll}\bar{\alpha} & \bar{\beta} & \bar{\gamma}\end{array}\right]=0$
- B
$\left[\begin{array}{lll}\bar{\alpha} & \bar{\beta} & \bar{\gamma}\end{array}\right]=1$
- C
$\left[\begin{array}{lll}\bar{\alpha} & \bar{\beta} & \bar{\gamma}\end{array}\right]=3$
- D
$\left[\begin{array}{lll}\bar{\beta} & \bar{\gamma} & \bar{\alpha}\end{array}\right]=1$
AnswerCorrect option: A. $\left[\begin{array}{lll}\bar{\alpha} & \bar{\beta} & \bar{\gamma}\end{array}\right]=0$
(A) Vector $\bar{\alpha}$ lies in the plane of $\bar{\beta}$ and $\bar{\gamma}$
$\begin{array}{ll}\therefore & \bar{\alpha}, \bar{\beta}, \bar{\gamma} \text { are coplanar. } \\ & \Rightarrow[\bar{\alpha} \bar{\beta} \bar{\gamma}]=0\end{array}$
View full question & answer→MCQ 2772 Marks
The scalar triple product of vectors is zero if __________
- ✓
One of the vector is zero vector
- B
Any two vectors are non-collinear
- C
the three vectors are non-coplanar
- D
AnswerCorrect option: A. One of the vector is zero vector
View full question & answer→MCQ 2782 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are three non-coplanar vectors, then $\frac{\bar{a} \cdot \bar{b} \times \bar{c}}{\bar{c} \times \bar{a} \cdot \bar{b}}+\frac{\bar{b} \cdot \bar{a} \times \bar{c}}{\bar{c} \cdot \bar{a} \times \bar{b}}=$
Answer(A) $\frac{\bar{a} \cdot \bar{b} \times \bar{c}}{\bar{c} \times \bar{a} \cdot \bar{b}}+\frac{\bar{b} \cdot \bar{a} \times \bar{c}}{\bar{c} \cdot \bar{a} \times \bar{b}}=\frac{\bar{a} \cdot \bar{b} \times \bar{c}}{\bar{c} \cdot \bar{a} \times \bar{b}}+\frac{\bar{b} \cdot \bar{a} \times \bar{c}}{\bar{c} \cdot \bar{a} \times \bar{b}}$
$=\frac{[\overline{ a } \overline{ b } \overline{ c }]}{[\overline{ c } \overline{ a } \overline{ b }]}+\frac{[\overline{ b } \overline{ a } \overline{ c }]}{[\overline{ c } \overline{ a } \overline{ b }]}$
$=\frac{[\overline{ a } \overline{ b } \overline{ c }]}{[\overline{ c } \overline{ a } \overline{ b }]}-\frac{[\overline{ a } \overline{ b } \overline{ c }]}{[\overline{ c } \overline{ a } \overline{ b }]}=0$
View full question & answer→MCQ 2792 Marks
$[\hat{ i } \hat{ k } \hat{ j }]+[\hat{ k } \hat{ j } \hat{ i }]+[\hat{ j } \hat{ k } \hat{ i }]=$
Answer(D) $[\hat{i} \hat{k} \hat{j}]+[\hat{k} \hat{j} \hat{i}]+[\hat{j} \hat{k} \hat{i}]=[\hat{i} \hat{k} \hat{j}]+[\hat{i} \hat{k} \hat{j}]-[\hat{i} \hat{k} \hat{j}]$
$\begin{array}{l}=[\hat{ i } \hat{ k } \hat{ j }] \\ =-1\end{array}$
View full question & answer→MCQ 2802 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are three vectors, then $[\bar{a} \quad \bar{b} \quad \bar{c}]$ is not equal to
- A
$\left[\begin{array}{lll}\bar{b} & \bar{c} & \bar{a}\end{array}\right]$
- B
$\left[\begin{array}{lll}\bar{c} & \bar{a} & \bar{b}\end{array}\right]$
- C
$-\left[\begin{array}{lll}\overline{ b } & \overline{ a } & \overline{ c }\end{array}\right]$
- ✓
$\left[\begin{array}{lll}\overline{ b } & \overline{ a } & \overline{ c }\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{lll}\overline{ b } & \overline{ a } & \overline{ c }\end{array}\right]$
(D) Since $[\overline{ a } \overline{ b } \overline{ c }]=[\overline{ b } \overline{ c } \overline{ a }]=[\overline{ c } \overline{ a } \overline{ b }]=-[\overline{ b } \overline{ a } \overline{ c }]$
View full question & answer→MCQ 2812 Marks
Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three vectors. Then scalar triple product $[\bar{a} \quad \bar{b} \quad \bar{c}]$ is equal to
- A
$\left[\begin{array}{lll}\bar{b} & \bar{a} & \bar{c}\end{array}\right]$
- B
$\left[\begin{array}{lll}\bar{a} & \bar{c} & \bar{b}\end{array}\right]$
- C
$\left[\begin{array}{lll}\bar{c} & \bar{b} & \bar{a}\end{array}\right]$
- ✓
$\left[\begin{array}{lll}\bar{b} & \bar{c} & \bar{a}\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{lll}\bar{b} & \bar{c} & \bar{a}\end{array}\right]$
View full question & answer→MCQ 2822 Marks
The value of $(\hat{i}+\hat{j}) \cdot[(\hat{j}+\hat{k}) \times(\hat{k}+\hat{i})]$ is
Answer(D) $(\hat{i}+\hat{j}) \cdot[(\hat{j}+\hat{k}) \times(\hat{k}+\hat{i})]$
$=\left|\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right|$
$=1(1)-1(-1)=2$
View full question & answer→MCQ 2832 Marks
The scalar triple product of the vectors $2 \hat{i}, 3 \hat{j}$ and $-5 \hat{k}$ is
Answer(D) $2 \hat{ i } \cdot[3 \hat{ j } \times(-5 \hat{ k })]=-30[\hat{ i } \cdot(\hat{ j } \times \hat{ k })]$
$\begin{array}{l}=-30(\hat{ i } \cdot \hat{ i })=-30(1) \\ =-30\end{array}$
View full question & answer→MCQ 2842 Marks
Which of the following expression are meaningful?
- ✓
$\overline{ u } \cdot(\overline{ v } \times \overline{ w })$
- B
$(\bar{u} \cdot \bar{v}) \cdot \bar{w}$
- C
$(\bar{u} \cdot \bar{v}) \times \bar{w}$
- D
$\bar{u} \times(\bar{v} \cdot \bar{w})$
AnswerCorrect option: A. $\overline{ u } \cdot(\overline{ v } \times \overline{ w })$
View full question & answer→MCQ 2852 Marks
If $\hat{i} ,\hat{j},\hat{k} $ are the unit vectors and mutually perpendicular, then [$\hat{i} ,\hat{k},\hat{j} $] is equal to
Answer(B) $[\hat{i} \hat{k} \hat{j}]=\hat{i} \cdot(\hat{k} \times \hat{j})=\hat{i} \cdot(-\hat{i})=-1$
View full question & answer→MCQ 2862 Marks
If $\bar{a}=2 \hat{i}+\hat{j}-\hat{k}, \quad \bar{b}=\hat{i}+2 \hat{j}+\hat{k} \quad$ and $\quad \bar{c}=\hat{i}-\hat{j}+2 \hat{k}$, then $\overline{ a } \cdot(\overline{ b } \times \overline{ c })=$
Answer(C) $\overline{ a } \cdot(\overline{ b } \times \overline{ c })=\left|\begin{array}{ccc}2 & 1 & -1 \\ 1 & 2 & 1 \\ 1 & -1 & 2\end{array}\right|$
$\begin{array}{l}=2(4+1)-1(2-1)-1(-1-2) \\ =12\end{array}$
View full question & answer→MCQ 2872 Marks
If the diagonals of a parallelogram are represented by the vectors $3 \hat{i}+\hat{j}-2 \hat{k}$ and $\hat{i}+3 \hat{j}-4 \hat{k}$, then its area is
- A
$5 \sqrt{3}$
- B
$6 \sqrt{3}$
- ✓
$\sqrt{42}$
- D
$\sqrt{28}$
AnswerCorrect option: C. $\sqrt{42}$
(C) Let $\bar{a}=3 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+3 \hat{j}-4 \hat{k}$ Then,
$\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 3 & -4\end{array}\right|=2 \hat{i}+10 \hat{j}+8 \hat{k}$
$\Rightarrow|\overline{ a } \times \overline{ b }|=\sqrt{4+100+64}=\sqrt{168}=2 \sqrt{42}$
$\therefore \quad$ Required area $=\frac{1}{2}|\overline{ a } \times \overline{ b }|=\sqrt{42}$ sq. units
View full question & answer→MCQ 2882 Marks
The area of the parallelogram whose adjacent sides are $\hat{i}-\hat{k}$ and $2 \hat{j}+3 \hat{k}$ is
- A
- B
- ✓
$\sqrt{17}$
- D
$2 \sqrt{13}$
AnswerCorrect option: C. $\sqrt{17}$
(C) $\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 0 & 2 & 3\end{array}\right|$
$=2 \hat{ i }-3 \hat{ j }+2 \hat{ k }$
∴ Area of parallelogram $=|\overline{ a } \times \overline{ b }|$
$\begin{array}{l}=\sqrt{4+9+4} \\ =\sqrt{17}\end{array}$
View full question & answer→MCQ 2892 Marks
Let $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of the vertices of a triangle ABC . The vector area of triangle ABC is
- A
$\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}$
- B
$\frac{1}{4}|\overline{ a } \times \overline{ b }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }|$
- ✓
$\frac{1}{2}|\overline{ a } \times \overline{ b }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }|$
- D
$\bar{b} \times \bar{a}+\bar{c} \times \bar{b}+\bar{a} \times \bar{c}$
AnswerCorrect option: C. $\frac{1}{2}|\overline{ a } \times \overline{ b }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }|$
View full question & answer→MCQ 2902 Marks
If $(\overline{ a } \times \overline{ b })^2+(\overline{ a } \cdot \overline{ b })^2=144$ and $|\overline{ a }|=4$, then $|\overline{ b }|=$
Answer(C) $(\overline{ a } \times \overline{ b })^2+(\overline{ a } \cdot \overline{ b })^2=|\overline{ a }|^2|\overline{b}|^2$
If $\bar{a}, \bar{b}$ are two vectors, then $(\bar{a} \times \bar{b})^2=\left|\begin{array}{ll}\bar{a} \cdot \bar{a} & \bar{a} \cdot \bar{b} \\ \bar{a} \cdot \bar{b} & \bar{b} \cdot \bar{b}\end{array}\right|$ or $|\overline{ a } \times \overline{ b }|^2+|\overline{ a } \cdot \overline{ b }|^2=|\overline{ a }|^2|\overline{b}|^2$ where $0<\theta<\frac{\pi^{ c }}{2}$
$\begin{array}{ll}\therefore & 144=16|\overline{b}|^2 \\ & \Rightarrow|\overline{b}|=3\end{array}$
View full question & answer→MCQ 2912 Marks
If $\theta$ is the angle between the vectors $\overline{ a }$ and $\overline{ b }$ and $|\overline{ a } \times \overline{ b }|=\overline{ a } \cdot \overline{ b }$, then $\theta$ is equal to
- A
$\pi$
- B
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
- D
$0$
AnswerCorrect option: C. $\frac{\pi}{4}$
(C) $|\overline{ a } \times \overline{ b }|=(\overline{ a } \cdot \overline{ b })$
$\Rightarrow| a || b | \sin \theta=| a || b | \cos \theta$
$\Rightarrow \tan \theta=1$
$\Rightarrow \theta=\frac{\pi}{4}$
View full question & answer→MCQ 2922 Marks
If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$, then $\frac{|\overline{ a } \times \overline{ b }|}{|\overline{ a } \cdot \overline{ b }|}$ is
- ✓
$\tan \theta$
- B
$-\tan \theta$
- C
$\cot \theta$
- D
$-\cot \theta$
AnswerCorrect option: A. $\tan \theta$
(A) Angle between the given vectors $\bar{a}$ and $\bar{b}$ is $\theta$ Since $\frac{|\overline{ a } \times \overline{ b }|}{|\overline{ a } \cdot \overline{ b }|}=\frac{|\overline{ a }||\overline{ b }| \sin \theta}{|\overline{ a }||\overline{ b }| \cos \theta}=\tan \theta$
View full question & answer→MCQ 2932 Marks
Let $\overline{ a }$ be a unit vector perpendicular to unit vectors $\bar{b}$ and $\bar{c}$ and if the angle between $\bar{b}$ and $\bar{c}$ is $\alpha$, then $\bar{b} \times \bar{c}$ is
- A
$\pm(\cos \alpha) \bar{a}$
- B
$\pm(\operatorname{cosec} \alpha) \bar{a}$
- ✓
$\pm(\sin \alpha) \overline{ a }$
- D
AnswerCorrect option: C. $\pm(\sin \alpha) \overline{ a }$
(C) Here, $\overline{ a }= \pm \frac{\overline{ b } \times \overline{ c }}{|\overline{ b } \times \overline{ c }|}$
$\Rightarrow \overline{ b } \times \overline{ c }= \pm|\overline{ b } \times \overline{ c }| \overline{ a }$
$= \pm(\sin \alpha) \overline{ a } \ldots[\because|\overline{ b } \times \overline{ c }|=\sin \alpha]$
View full question & answer→MCQ 2942 Marks
If $|\bar{a} \times \bar{b}|=|\bar{a}||\bar{b}|$, then $\bar{a}$ and $\bar{b}$ are
View full question & answer→MCQ 2952 Marks
If cross product of two non-zero vectors is zero, then the vectors are
MCQ 2962 Marks
If $\bar{a}=\hat{i}+\hat{j}-\hat{k}, \quad \bar{b}=-\hat{i}+2 \hat{j}+2 \hat{k} \quad$ and $\overline{ c }=-\hat{ i }+2 \hat{ j }-\hat{ k }$, then a unit vector perpendicular to the vectors $\bar{a}+\bar{b}$ and $\bar{b}-\bar{c}$ is
- A
$\hat{ i }$
- B
$\hat{ j }$
- C
$\hat{ k }$
- D
View full question & answer→MCQ 2972 Marks
A unit vector perpendicular to the plane of $\bar{a}=2 \hat{i}-6 \hat{j}-3 \hat{k}$ and $\bar{b}=4 \hat{i}+3 \hat{j}-\hat{k}$ is
- A
$\frac{1}{\sqrt{26}}(4 \hat{ i }+3 \hat{ j }-\hat{ k })$
- B
$\frac{1}{7}(2 \hat{ i }-6 \hat{ j }-3 \hat{ k })$
- C
$\frac{1}{7}(3 \hat{ i }-2 \hat{ j }+6 \hat{ k })$
- D
$\frac{1}{7}(2 \hat{ i }-3 \hat{ j }-6 \hat{ k })$
View full question & answer→MCQ 2982 Marks
A unit vector perpendicular to the plane containing the vectors $\hat{i}-\hat{j}+\hat{k}$ and $-\hat{ i }+\hat{ j }+\hat{ k }$ is
- A
$\frac{\hat{ i }-\hat{ j }}{\sqrt{2}}$
- B
$\frac{\hat{ i }-\hat{ k }}{\sqrt{2}}$
- C
$\frac{\hat{ j }-\hat{ k }}{\sqrt{2}}$
- D
$\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
View full question & answer→MCQ 2992 Marks
If $\bar{a}=3 \hat{i}-5 \hat{j}$ and $\bar{b}=6 \hat{i}+3 \hat{j}$ are two vectors and $\overline{ c }$ is a vector such that $\overline{ c }=\overline{ a } \times \overline{ b }$, then $|\overline{ a }|:|\overline{ b }|:|\overline{ c }|=$
View full question & answer→MCQ 3002 Marks
$(2 a+3 b) \times(5 a+7 b)=$
- A
$a \times b$
- B
$b \times a$
- C
$a+b$
- D
$7 a+10 b$
View full question & answer→