Vectors & 3D Geometry PYQ — Page 19

This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1 : The point $A(1,0,7)$ is the mirror image of the point $B(1,6,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. Statement-2 : The line: $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ bisects the line segment joining $A(1,0,7)$ and $B(1,6,3)$.

A line $A B$ in three-dimensional space makes angles $45^{\circ}$ and $120^{\circ}$ with the positive $x$-axis and the positive $y$-axis respectively. If $A B$ makes an acute angle $\theta$ with the positive $z$-axis, then $\theta$ equals

Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$. Then vector $\vec{b}$ satisfying $\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}$ and $\vec{a} \cdot \vec{b}=3$ is

If the vectors $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mu \hat{\mathrm{k}}$ are mutually orthogonal, then $(\lambda, \mu)=$

The projections of a vector on the three coordinate axis are $6,-3,2$ respectively. The direction cosines of the vector are

The non-zero verctors $\vec{a}, \vec{b}$ and $\vec{c}$ are related by $\vec{a}=8 \vec{b}$ and $\vec{c}=-7 \vec{b}$. Then the angle between $\vec{a}$ and $\overrightarrow{\mathrm{c}}$ is

The vector $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\hat{j}+\hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$ ?

If the straight lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect at a point, then the integer $k$ is equal to

The line passing through the points $(5,1, a)$ and $(3, b, 1)$ crosses the $y z-$ plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then

Let $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{x} \hat{\mathrm{i}}+(\mathrm{x}-2) \hat{\mathrm{j}}-\hat{\mathrm{k}}$. If the vector $\overline{\mathrm{c}}$ lies in the plane of $\bar{a}$ and $\bar{b}$, then $x$ equals

The resultant of two forces $\mathrm{P~N}$ and $3 \mathrm{~N}$ is a force of $7 \mathrm{~N}$. If the direction of $3 \mathrm{~N}$ force were reversed, the resultant would be $\sqrt{19} \mathrm{~N}$. The value of $\mathrm{P}$ is

If a line makes an angle of $\frac{\pi}{4}$ with the positive directions of each of $x$-axis and $y$-axis, then the angle that the line makes with the positive direction of the $z-$axis is

If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them, then $2 \hat{u} \times 3 \hat{v}$ is a unit vector for

A particle has two velocities of equal magnitude inclined to each other at an angle $\theta$. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $\theta$ is

The values of $a$, for which the points $A, B, C$ with position vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $a \hat{i}-3 \hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $C=\frac{\pi}{2}$ are

If $(\mathrm{a} \times \mathrm{b}) \times \overline{\mathrm{c}}=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$, where $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are any three vectors such that $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \neq 0$, $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \neq 0$, then $\mathrm{a}$ and $\mathrm{c}$ are

$A B C$ is a triangle, right angled at $A$. The resultant of the forces acting along $\overrightarrow{A B}, \overrightarrow{A C}$ with magnitudes $\frac{1}{A B}$ and $\frac{1}{A C}$ respectively is the force along $\overrightarrow{A D}$, where $D$ is the foot of the perpendicular from $\mathrm{A}$ onto $\mathrm{BC}$. The magnitude of the resultant is

The line parallel to the $x$-axis and passing through the intersection of the lines ax $+$ $2 b y+3 b=0$ and $b x-2 a y-3 a=0$, where $(a, b) \neq(0,0)$ is

The resultant $R$ of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to smaller one is

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number then $\left[\lambda(\vec{a}+\vec{b}) \lambda^2 \vec{b} \lambda \vec{c}\right]=[\vec{a} \vec{b}+\vec{c} \vec{b}]$ for

For any vectora a the value of $(\vec{a} \times \hat{i})^2+(\vec{a} \times \hat{j})^2+(\vec{a} \times \hat{k})^2$ is equal to

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ lie in a plane, then $c$ is

Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$. Then $[\vec{a}, \vec{b}, \vec{c}]$ depends on

If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$, then