3D Geometry Questions (578)

The distance of the plane $x + 2y - z = 2$ from the point $(2, -1, 3)$, measured in the direction with the direction ratios $(2, 2, 1)$ is

The plane $3x + 4y + 12z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. If the equation of the plane in new position is $x - 4y + 6x = K$, then the value of $K$ is

The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is

Let $\vec{a}$, $\vec{b}$ be two vectors perpendicular to each other with $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{c}\times\vec{a}=\vec{b}$. The least value of $|\vec{c}-\vec{a}|$ is

Consider a tetrahedron D—ABC with position vectors of its angular points as A(1, 1, 1); B(1, 2, 3); C(1, 1, 2) and centre of tetrahedron \left(\frac{3}{2}, \frac{3}{4}, 2\right). Find the shortest distance between the skew lines AB and CD.

Let \(P\) be the plane, which contains the line of intersection of the planes, \(x + y + z - 6 = 0\) and \(2x + 3y + z + 5 = 0\) and it is perpendicular to the \(xy\)-plane. Then the distance of the point \((0, 0, 256)\) from \(P\) is ______ (up to three decimal places).

The position vectors of the points P and Q are \(5\hat{i} + 7\hat{j} - 2\hat{k}\) and \(-3\hat{i} + 3\hat{j} + 6\hat{k}\), respectively. The vector \(\vec{A} = 3\hat{i} - \hat{j} + \hat{k}\) passes through the point P and the vector \(\vec{B} = -3\hat{i} + 2\hat{j} + 4\hat{k}\) passes through the point Q. A third vector \(2\hat{i} + 7\hat{j} - 5\hat{k}\) intersects vectors \(\vec{A}\) and \(\vec{B}\). Then the distance between the two position vectors of the points of intersection is __________ (up to three decimal places).

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $P$, be $(\alpha,\beta,\gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to:

The number of planes that are equidistant from four non-coplanar points is

The shortest distance between the lines $\dfrac{x-3}{4}=\dfrac{y+7}{-11}=\dfrac{z-1}{5}$ and $\dfrac{x-5}{3}=\dfrac{y-9}{-6}=\dfrac{z+2}{1}$ is:

A line makes an angle $\theta$ both with X and Y-axes. A possible value of $\theta$ is in

The shortest distance between the line $x = y = z$ and the line of intersection of $2x + y + z - 1 = 0$ and $3x + y + 2z - 2 = 0$ is

Find the value of $\lambda$ if the plane $2x - 2y + z - 3 + \lambda = 0$ contains the point $(3, 1, 1)$.

Let $L_1$, $L_2$ be two distinct lines such that $L_1$ and $L_2$ are not perpendicular. Let $P$ be a plane such that $L_1$ and $L_2$ are not perpendicular to $P$. Let $L_3$ and $L_4$ be projections of $L_1$ and $L_2$ on plane $P$ respectively. If $\theta$ be angle between $L_3$ and $L_4$ then $\theta$ CANNOT be equal to

If the equation of the plane through $(-1,2,0)$ and parallel to lines $\dfrac{x}{3}=\dfrac{y+1}{0}=\dfrac{z-2}{-1}$ and $\dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z+1}{-1}$ is $ax+by+cz=1$, then $a+b+c$ is

Consider lines $L_1:\{3\sqrt3x=4\sin\theta\,y+(2\sqrt{\sin\theta}-3)\,3\sqrt3,\;3\sqrt3z=(2\sqrt{\sin\theta}-3)y+3\sqrt3\sin\theta\}$ and $L_2:\{\sqrt3x=-2\sqrt{\cos\theta}\,y+\sqrt3(3-2\sqrt{\cos\theta}),\;\sqrt3z=(3-2\sqrt{\cos\theta})y+\sqrt3\cos\theta\}$. If $L_1\perp L_2$ then

The distance of the point $(-1, 9, -16)$ from the plane $2x + 3y - z = 5$ measured parallel to the line $\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$ is:

Consider the lines $L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$ and $L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$. A line $L_3$ having direction ratios $1, -1, -2$, intersects $L_1$ and $L_2$ at the points P and Q respectively. Then the length of line segment PQ is

If the shortest distance between the lines $\frac{x+\sqrt{6}}{2} = \frac{y-\sqrt{6}}{3} = \frac{z-\sqrt{6}}{4}$ and $\frac{x-\lambda}{3} = \frac{y-2\sqrt{6}}{4} = \frac{z+2\sqrt{6}}{5}$ is 6, then the square of sum of all possible values of $\lambda$ is

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r}\cdot(3\hat{i}-5\hat{j}+\hat{k}) = 7$ and $P_2: \vec{r}\cdot(\lambda\hat{i}+\hat{j}-3\hat{k}) = 9$ is $\sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)$, then the square of the length of perpendicular from the point $(38\lambda_1, 10\lambda_2, 2)$ to the plane $P_1$ is _____.

Let a line L pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of L from the point $(5, 3, 8)$ is $\alpha$, then $3\alpha^2$ is equal to _____.

Let the shortest distance between the lines $L: \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$, $\lambda \geq 0$ and $L_1: x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on L, then which of the following is NOT possible?

Let the image of the point P(2, -1, 3) in the plane $x + 2y - z = 0$ be Q. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point Q is

Let $\alpha x + \beta y + yz = 1$ be the equation of a plane passing through the point $(3, -2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 5)$. Then the value of $\alpha\beta y$ is equal to _____.

Let $(\alpha,\beta,\gamma)$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}$. Then $19(\alpha+\beta+\gamma)$ is equal to:

Let $\vec{a}=3\hat i-4\hat j+5\hat k$, $\vec{b}=\hat i+2\hat j-2\hat k$, $\vec{c}=5\hat i-12\hat j-13\hat k$. If $\vec{r}\times\vec{b}=\vec{b}\times(\vec{a}-\vec{c})$ and $\vec{r}\cdot(\vec{a}+\vec{c})=0$, then $|\vec{r}-36\vec{b}|^2$ is

If the length of the perpendicular from point $P(a, 4, 2)$, $a>0$, to the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}$ is $2\sqrt{6}$ units, and $Q(\alpha_1,\alpha_2,\alpha_3)$ is the image of $P$ on this line, then $a+\displaystyle\sum_{i=1}^3 \alpha_i$ equals

Let $P$ be the foot of the perpendicular from the point $M(1,2,2)$ on the line $L:\dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z-2}{2}$. Let the line $\overrightarrow{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$, $\lambda\in\mathbf{R}$, intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to:

If the line $\dfrac{2-x}{3}=\dfrac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\dfrac{x+3}{3\mu}=\dfrac{1-2y}{6}=\dfrac{5-z}{7}$, then $4\lambda+9\mu$ is equal to:

Let $P(x,y,z)$ be a point in the first octant, whose projection in the $xy$-plane is the point $Q$. Let $OP=\gamma$; the angle between $OQ$ and the positive $x$-axis be $\theta$; and the angle between $OP$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is

Let the line $L$ intersect the lines $x-2=-y=z-1$, $2(x+1)=2(y-1)=z+1$ and be parallel to the line $\dfrac{x-2}{3}=\dfrac{y+1}{1}=\dfrac{z-2}{2}$. Then which of the following points lies on $L$?

One vertex of a rectangular parallelopiped is at origin $O$; edges along axes are 3, 4, 5. $P=(3,4,5)$. Shortest distance between diagonal $OP$ and an edge parallel to $z$-axis (not through $O$ or $P$) is

Plane through intersection of $2x-y+z=3$ and $4x-3y+5z+9=0$, parallel to $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$, is $ax+by+cz+6=0$. Then $a+b+c$ is

Shortest distance between $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ is

$O$ origin, $P=-\hat{i}-2\hat{j}+3\hat{k}$. $A=-2\hat{i}+\hat{j}-3\hat{k}$, $B=2\hat{i}+4\hat{j}-2\hat{k}$, $C=-4\hat{i}+2\hat{j}-\hat{k}$. Projection of $\overrightarrow{OP}$ on vector $\perp\overrightarrow{AB}$ and $\overrightarrow{AC}$ is

Line $L:\ x=\frac{1-y}{-2}=\frac{z-3}{\lambda}$ meets plane $x+2y+3z=4$ at $(\alpha,\beta,\gamma)$. Angle between $L$ and plane is $\cos^{-1}(\sqrt{\frac{5}{14}})$. Then $\alpha+2\beta+6\gamma$ is equal to

Image of $(5,5,8)$ in $x-2y+z-2=0$ is $P$. Distance of $Q(6,-2,\alpha)$, $\alpha>0$ from $P$ is $\frac{13}{3}$. Then $\alpha$ is equal to _______

$S$ = set of $\lambda$ for which SD between $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8|\sum_{\lambda\in S}\lambda|$ is equal to

Foot of $\perp$ of $P(3,-2,-9)$ on plane through $(-1,-2,-3),(9,3,4),(9,-2,1)$ is $Q(\alpha,\beta,\gamma)$. Distance of $Q$ from origin is

Plane $P$ contains $2x+y-z-3=0=5x-3y+4z+9$ and is parallel to $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Distance of $A(8,-1,-19)$ from $P$ measured $\parallel$ to $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to _________

Line $x=y=z$ intersects $x\sin A+y\sin B+z\sin C-18=0=x\sin2A+y\sin2B+z\sin2C-9$ where $A,B,C$ are angles of $\triangle ABC$. Then $80\!\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$ is equal to _________

The equation of a line is \(\dfrac{x}{1} = \dfrac{y-1}{0} = \dfrac{z+1}{-1}\). From point \(P(\beta, 0, \beta)\) where \(\beta \neq 0\), a perpendicular is drawn to the given line meeting at point \(M(\lambda, 1, -\lambda-1)\). Find the value of \(\beta\).

If the image of the point (4, 4, 3) in the line x-1 = y-2 = z-1 is (\alpha, \beta, \gamma), then \alpha + \beta + \gamma is equal to 2 1 3

The perpendicular distance of a corner of a unit cube from a diagonal not passing through it is

Let the line of the shortest distance between the lines $L_1:\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $L_2:\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha,\beta,\gamma)$ is the midpoint of the line segment $PQ$, then $2(\alpha+\beta+\gamma)$ is equal to

Let $P$ and $Q$ be the points on the line $\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}$ which are at a distance of 6 units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha,\beta,\gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is:

The perpendicular distance, of the line y+2 x-1 2 = -1 = z+3 2 from the point P(2, -10, 1) , is :

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle P QR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR) is equal to _______ -. 2

If the shortest distance between the lines $\dfrac{x-4}{1}=\dfrac{y+1}{2}=\dfrac{z}{-3}$ and $\dfrac{x-\lambda}{2}=\dfrac{y+1}{4}=\dfrac{z-2}{-5}$ is $\dfrac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is:

Let the image of the point $(1,0,7)$ in the line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$ be the point $(\alpha,\beta,\gamma)$. Then which one of the following points lies on the line passing through $(\alpha,\beta,\gamma)$ and making angles $\dfrac{2\pi}{3}$ and $\dfrac{3\pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with $x$-axis?