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If the lines $x = ay+b$, $z = cy+d$ and $x = py+q$, $z = ry+s$ are perpendicular, then $ap + cr$ is equal to

A line is perpendicular to two lines having direction ratios $1,-2,-2$ and $0,2,1$ then the direction cosines of the line are:

The distance between the two lines $L_1$ and $L_2$ given by $L_1: \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{-z-4}{-6}$ and $L_2: \dfrac{x-3}{2} = \dfrac{y-3}{3} = \dfrac{z+5}{6}$ is

The shortest distance between lines whose vector equations are: $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$, is:

The equation of line passing through the points (2, -1, 3) and (4, 2, 1) is

(A) $\frac{x-2}{2} = \frac{y+1}{3} = \frac{z-3}{-2}$

(B) $\frac{x-4}{3} = \frac{y-2}{-2} = \frac{z-1}{2}$

(C) $\vec{r} = (2\hat{i}-\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}-2\hat{k})$

(D) $\frac{x-4}{-2} = \frac{y-2}{-3} = \frac{z-1}{2}$

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Direction ratios of the line perpendicular to the lines $\frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1}$ and $\frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2}$ are

Given a line $L:\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$, then

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The shortest distance between lines $L_1:\frac{x+3}{-4}=\frac{6-y}{-3}=\frac{z}{2}$ and $L_2:\frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1}$ is:

The Cartesian equation of a line is $6x-2=3y+1=2z-2$. Then its vector from is:

Let $L_1$ and $L_2$ lines such that $L_1:\vec{r}=\hat{i}+\hat{j}+\lambda(2\hat{i}-\hat{j}+\hat{k})$, $L_2:\vec{r}=2\hat{i}+\hat{j}-\hat{k}+\mu(4\hat{i}-2\hat{j}+2\hat{k})$ and $\theta$ be the acute angle between $L_1$ and $L_2$, then

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The perpendicular distance of point (1, 1, 3) on the line $\frac{x}{2}=\frac{y+1}{3}=\frac{-z-3}{-5}$ is

If $\lambda$ and $\mu$ are parameters, then

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P is a point on the line segment joining the points (3, 2, -1) and (6, 2, -2) internally. If x coordinate of P is 5, then its z-coordinate is

If the shortest distance between the lines $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$ and $\vec r=(a\hat i+4\hat j+5\hat k)+\mu(4\hat i+6\hat j+8\hat k)$, $a \in N$, where $\lambda$ and $\mu$ are parameters, is $\sqrt{\dfrac{5}{29}}$, then $a$ is equal to

Which of the following statements are correct?

(A) The angle between the vectors with direction ratios proportional to -3, 5, 4 and 4, 5, 3 is $\dfrac{\pi}{3}$

(B) The direction cosines of Y-axis is 0, 1, 0.

(C) The cartesian equation of a line is $3x-6=2y+1=2z-1$, then its direction ratios are proportional to 2, 2, 3

(D) The vector equation of the line passing through the point $(2,-1,-1)$ and direction ratios 1, 2, 3 is $\vec r=(2\hat i-\hat j-\hat k)+\lambda(\hat i+2\hat j+3\hat k)$

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A line in space is uniquely determined if

(A) It passes through a given point and has a given direction.

(B) It passes through two given points.

(C) It passes through a given point.

(D) It has a given direction.

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If the points $P(3,\mu,6), Q(-1,-1,2)$ and $R(2,8,5)$ are collinear, then the value of '$\mu$' is

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The value of $\lambda$ so that the lines $\dfrac{1-x}{3} = \dfrac{7y-14}{2\lambda} = \dfrac{z-3}{2}$ and $\dfrac{7-7x}{3\lambda} = \dfrac{y-5}{1} = \dfrac{6-z}{5}$

are at right angles is:

If shortest distance between the lines $\vec{r} = \hat{i} - \hat{j} + \lambda(2\hat{i}+\hat{k})$ and

$\vec{r} = 2\hat{i} - \hat{j} + \mu(\hat{i}-\hat{j}-\hat{k})$ is $\dfrac{\sqrt{14}}{14}m$ units, the value of $m$ is

Two points are moving along the lines :

$L_1: \vec{r} = (2\hat{i}+3\hat{j}+\hat{k}) + \lambda(\hat{i}+2\hat{j}+\hat{k})$

$L_2: \vec{r} = (\hat{i}+\hat{j}) + \mu(2\hat{i}+\hat{j}-\hat{k})$

where $\lambda$ and $\mu$ are real parameters , then which of the following statements are correct ?

A. The direction ratios of the line perpendicular to both lines are $<-1,1,-1>$.

B. The direction cosines of the second line $L_2$ are $<\dfrac{2}{\sqrt6},\dfrac{-1}{\sqrt6},\dfrac{-1}{\sqrt6}>$.

C. Angle between two lines is $\pi/3$.

D. The lines are skew lines.

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The cartesian equation of the line passing through the point $(-4, 3, -2)$ and parallel to the line $\vec{r} = 5\hat{i} - \hat{k} + \lambda(\hat{i}-\hat{j}+2\hat{k})$ is:

The angle between the lines $\dfrac{3-x}{-7} = \dfrac{y+5}{-5} = \dfrac{z}{1}$ and $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ is:

The points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ at a distance of $\sqrt{17}$ from the point (-2, -1, 3) are:

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The shortest distance (in units) between the lines $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $L_2:\frac{x-3}{2}=\frac{3-y}{-3}=\frac{z+5}{6}$ is:

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Direction cosines of 2 lines are given by the equations $3l+m+5n=0, 6mn-2nl+5lm=0$ then which of the following statements are correct ?

(A) Direction ratios of two lines are (1, 2, -1) and (-2, 1, 1).

(B) Angle between lines is $\frac{\pi}{3}$.

(C) The angle between the lines is $\cos^{-1}\left(-\frac{1}{6}\right)$.

(D) Vectors parallel to these lines are $\hat i+2\hat j-\hat k$ and $-2\hat i+\hat j+\hat k$.

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If the lines $\frac{1-x}{2}=\frac{2y}{p}=\frac{z-1}{1}$ and $\frac{3-2x}{p}=\frac{y-1}{2}=\frac{z}{1}$ are parallel then value of 'p' is

Let ABCD be a parallelogram with vertices $A(2,2,a), B(1,0,0), C(1,3,a)$ and D, where $a > 0$. If the area of parallelogram is 5 square units, then the value of $a$ is: