Three Dimensional Geometry

The equation of the line passing through the point of intersection of $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and also through the point ( $2,1,-2$ ) is

Let the plane passing through point $(2,1,-1)$ containing line joining the points $(1,3,2)$ and $(1,2,1)$ makes intercepts $\mathrm{p}, \mathrm{q}, \mathrm{r}$ on co-ordinate axes, then $\mathrm{p}+\mathrm{q}+\mathrm{r}=$

The line passing through the points $(a, 1,6)$ and $(3,4, \mathrm{~b})$ crosses the $y z$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$, then the value of $(3 a+4 b)$ is

If $\theta$ is the angle between the lines whose direction cosines are given by $6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{~m}=0$ and $3 l+\mathrm{m}+5 \mathrm{n}=0$, then $\sin \theta=$

The angle between the lines $x=y, z=0$ and $y=0, \mathrm{z}=0$ is

If the foot of the perpendicular drawn from the origin to a plane is $\mathrm{P}(2,-1,4)$, then the equation of the plane is

The angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=6$ is $\cos ^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is

If the sum of the squares of the distances of a point $\mathrm{P}(x, y, z)$ from the three co-ordinate axes is 324 , then the distance of point P from the origin is ….

The perimeter of a square whose two sides have equations $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ is

The angle between the lines $3 x=2 y=-\mathrm{z}$ and $-x=6 y=-4 z$ is

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3 y-\alpha z+\beta=0$, then the value of $(\beta-\alpha)$ is equal to

If the planes $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\lambda \hat{\mathrm{j}}+\hat{\mathrm{k}})=3$ and $\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\mu \hat{\mathrm{k}})=5$ are parallel, then $\lambda+\mu=$

If the lines $\frac{x-1}{2}=\frac{y+1}{\mathrm{k}}=\frac{\mathrm{z}}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{\mathrm{z}}{\mathrm{k}}$ are coplanar, then the equation of the plane containing these lines are

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, \mathrm{z}=1$, then $\alpha+\beta=$

The angle between the lines $x-3 y-4=0,4 y-z+5=0$ and $x+3 y-11=0,2 y-z+6=0$ is

If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is

The distance of the plane $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ from the origin is

The line L is passing through $(1,2,3)$. The distance of any point on the line L from the line $\overline{\mathrm{r}}=(3 \lambda-1) \hat{\mathrm{i}}+(-2 \lambda+3) \hat{\mathrm{j}}+(4+\lambda) \hat{\mathrm{k}}$ is constant. Then the line L ...

If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha>0$ is 5 units, then the foot of the perpendicular from P to the plane is

The shortest distance between the lines $\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})$ is

The equation of the plane containing the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-4}{-2}$ and the point $(0,5,0)$ is

The angle between the lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{\mathrm{~m}}=\frac{y-3}{\mathrm{n}}=\frac{\mathrm{z}-1}{l}$, where $l>\mathrm{m}>\mathrm{n}$ and $1, \mathrm{~m}, \mathrm{n}$ are roots of the equation $x...

The angle between the lines whose direction cosines are $\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}$ and $\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}$ is

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3 y-6=0$, are

The distance of the point $\mathrm{P}(3,8,2)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3 x+2 y-2 z+15=0$ is

The length of the altitude through the point $D$ of tetrahedron where the vertices of the tetrahedron are $A(2,3,1), B(4,1,-2), C(6,3,7), D(-5,-4,8)$, is

The distance between the line $\overline{\mathrm{r}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and the plane $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+\hat{\mat...

The acute angle between the lines $x=-2+2 \mathrm{t}, y=3-4 \mathrm{t}, \mathrm{z}=-4+\mathrm{t}$ and $x=-2-\mathrm{t}, y=3+2 \mathrm{t}, \mathrm{z}=-4+3 \mathrm{t}$ is

If the angle between the planes $x-2 y+3 z-5=0$ and $x+\alpha y+2 z+7=0$ is $\cos ^{-1}\left(\frac{1}{14}\right)$ then the difference between the values of $\alpha$ is

The direction cosines of the line $x-y+2 z=5$ and $3 x+y+z=6$ are

If the shortest distance between the lines $\frac{x-\mathrm{k}}{2}=\frac{y-4}{3}=\frac{\mathrm{z}-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{\mathrm{z}-7}{8}$ is $\frac{13}{\sqrt{29}}$, then $\mathrm{k}=$

If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the co-ordinate axes at points $\mathrm{A}, \mathrm{B}$ and C , then the area of the triangle ABC is

The distance of the point $(2,4,0)$ from the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ is

The equation of the plane passing through the point of intersection of the planes $2 x-y+z-3=0$ and $4 x-3 y+5 z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$ Then $\alpha+\beta...

If the line $\frac{x+1}{3}=\frac{y-k}{7}=\frac{z-4}{8}$ lies in the plane $2 x+\mathrm{p} y+7 z-41=0$ which is perpendicular to the plane $x+4 y-2 z+13=0$ then $\mathrm{k}=$

The co-ordinates of the point where the line joining the points $(2,-3,1)$ and $(3,-4,-5)$ and intersects the plane $2 x+y+z=7$ are

A plane passes through $(1,-2,1)$ and is perpendicular to the planes $2 x-2 y+z=0$ and $x-y+2 z=4$. The distance of the point $(1,2,2)$ from this plane is ________ units.

The Cartesian equation of plane through $\mathrm{A}(7,8,6)$ and parallel to the XY plane is

If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=0$ is such that $\sin \theta=\frac{1}{3}$, then $\lambda+1=$

If the points $\mathrm{A}(2-x, 2,2), \mathrm{B}(2,2-y, 2)$, $\mathrm{C}(2,2,2-\mathrm{z})$ and $\mathrm{D}(1,1,1)$ are coplanar, then the locus of point $\mathrm{P}(x, y, \mathrm{z})$ is

If the directed line makes an angle $45^{\circ}$ and $60^{\circ}$ with the X and Y -axes respectively, then the obtuse angle $\theta$ made by the line with the Z -axis is

If the sum of the squares of the distance of the point $\mathrm{P}(x, y, \mathrm{z})$ from the co-ordinate axes is 242 , then the distance of the point P from the origin is units.

If the lines $\frac{3-x}{2}=\frac{5 y-2}{3 \lambda+1}=5-\mathrm{z}$ and $\frac{x+2}{-1}=\frac{1-3 y}{7}=\frac{4-z}{2 \mu}$ are at right angles, then $7 \lambda-10 \mu=$

The distance of the point $(-3,2,3)$ from the line passing through $(4,6,-2)$ and having direction ratios $-1,2,3$ is $\qquad$ units.

The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{\mathrm{k}}=\frac{y-3}{1}=\frac{\mathrm{z}-3}{2}$ are coplanar, hence $\mathrm{k}=$

If the shortest distance between the lines $\bar{r}_1=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in \mathbb{R}, \alpha>0 \quad$ and $\bar{r}_2=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu...

The coordinates of the foot of the perpendicular drawn from a point $\mathrm{P}(-1,1,2)$ to the plane $2 x-3 y+z-11=0$ are

A plane passes through $(2,1,2)$ and $(1,2,1)$ and parallel to the line $2 x=3 y$ and $\mathrm{z}=1$, then the plane also passes through the point

The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3 x+4 y+5 z=2$ and perpendicular to the XY- plane is

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