3D Geometry Questions

The two lines $x=ay+b,z=cy+d$; and $x=a^{\prime }y+b^{\prime },z=c^{\prime }y+d^{\prime }$ are perpendicular to each other if:

If the straight lines $x=1+s,y=-3-\lambda s,z=1+\lambda s$ and $x=\frac{t}{2},y=1+t,z=2-t$ with parameters $s$ and $t$ respectively, are co-planar then $\lambda$ equals

A line makes the same angle $\theta$ with each of the $x$ and $z$ axes. If the angle $\beta ,$ which it makes with $y$ -axis, is such that ${\sin }^2\beta =3{\sin }^2\theta ,$ then ${\cos }^2\theta$ equals

A line with direction cosines proportional to $2,1,2$ meets each of the lines $x=y+a=z$ and $x+a=2y=2z.$ The co-ordinates of each of the point of intersection are given by

Distance between two parallel planes $2x+y+2z=8$ and $4x+2y+4z+5=0$ is

The intersection of the spheres $x^2+y^2+z^2+7x-2y-z=13$ and $x^2+y^2+z^2-3x+3y+4z=8$ is the same as the intersection of one of the sphere and the plane

The image of the point $(-1,3,4)$ in the plane $x-2y=0$ is:

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 $yz$ -plane at the point $\left(0,\frac{17}{2},-\frac{13}{2}\right)$. Then:

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k$ is equal to:

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta =0$. Then $(\alpha ,\beta )$ equals:

Statement-1: The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5$.Statement-2: The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4)$.

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

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

Statement-1 : The point $A(1,0,7)$ is the mirror image of the point $B(1,6,3)$ in the line $\frac{x-1}{0}=\frac{y-1}{2}=\frac{z-2}{3}$.Statement-2 : The line $\frac{x-1}{0}=\frac{y-1}{2}=\frac{z-2}{3}$ bisects the line segment joining $A(1,0,7)$ and $B(1,6,3)$.

An equation of a plane parallel to the plane $x-2y+2z-5=0$ and at a unit distance from the origin is:

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

Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a,b,c$ and $a^{\prime },b^{\prime },c^{\prime }$ from the origin, then:

The shortest distance from the plane $12x+4y+3z=327$ to the sphere $x^2+y^2+z^2+4x-2y-6z=155$ is:

The two lines $x=ay+b$, $z=cy+d$ and $x=a^{\prime }y+b^{\prime }$, $z=c^{\prime }y+d^{\prime }$ will be perpendicular, if and only if:

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

The radius of the circle in which the sphere $x^2+y^2+z^2+2x-2y-4z-19=0$ is cut by the plane $x+2y+2z+7=0$ is:

A tetrahedron has vertices at $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$ and $C(-1,1,2)$. Then the angle between the faces $OAB$ and $ABC$ will be:

The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius:

The distance between the line $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+4\hat{k})$ and the plane $\vec{r}\cdot (\hat{i}+5\hat{j}+\hat{k})=5$ is:

If the plane $2ax-3ay+4az+6=0$ passes through the midpoint of the line joining the centres of the spheres $x^2+y^2+z^2+6x-8y-2z=13$ and $x^2+y^2+z^2-10x+4y-2z=8$, then $a$ equals:

The angle between the lines $2x=3y=-z$ and $6x=-y=-4z$ is: