3D Geometry Questions (578)

Let $PQR$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the midpoint of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the lines $\dfrac{x-2}{0}=\dfrac{y}{2}=\dfrac{z+3}{-1}$ and $\dfrac{x-1}{1}=\dfrac{y+3}{-3}=\dfrac{z+1}{1}$ is

$L$ through origin $\perp L_1$ and $L_2$. $P$ = intersection of $L$ and $L_1$. $Q(\alpha,\beta,\gamma)$ = foot of $\perp$ from $P$ on $L_2$. Then $9(\alpha+\beta+\gamma)$ is equal to ________.

The distance, of the point $(7,-2,11)$ from the line $\dfrac{x-6}{1}=\dfrac{y-4}{0}=\dfrac{z-8}{-3}$ along the line $\dfrac{x-5}{2}=\dfrac{y-1}{-3}=\dfrac{z-5}{6}$, is:

Plane through $(0,-1,2)$ and $(-1,2,1)$, parallel to line through $(5,1,-7)$ and $(1,-1,-1)$. Which point lies on it?

A line with direction ratios $2,1,2$ meets the lines $x=y+2=z$ and $x+2=2y=2z$ respectively at the points $P$ and $Q$. If the length of the perpendicular from the point $(1,2,12)$ to the line $PQ$ is $l$, then $l^2$ is

The shortest distance between the lines $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}$ and $\frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$ is equal to _______.

Plane $P$ contains line of intersection of $\vec{r}\cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r}\cdot(2\hat{i}+3\hat{j}+4\hat{k})=-5$. $P$ passes through $(0,2,-2)$. Square of distance from $(12,12,18)$ to $P$ is

A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60°$, to the y-axis at $45°$ and to the z-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$, is normal to $\vec{v}$, then

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

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{-2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{3}=\dfrac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ is:

Normal vector to the plane $P = 0$ is $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 2 \\ 3 & 6 & -2 \end{vmatrix} = \vec{i}(-10) - \vec{j}(-10) + \vec{k}(15) = -10\vec{i} + 10\vec{j} + 15\vec{k}$. Equation of the plane through $(1, 1, 1)$ is $-2(x - 1) + 2(y - 1) + 3(z - 1) = 0$. So, the distance from the point $(1, 2, 3)$ is $\left|\frac{-1 + 3 - 3}{\sqrt{17}}\right| = \frac{1}{\sqrt{17}}$ units.

Let L : y-2 y-4 and L : be two lines. Then which of the following points lies on x-1 z-3 x-2 z-5 1 = = 2 = = 2 3 4 3 4 5 the line of the shortest distance between L and L ? 1 2

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x + 2y + z = 0$ and $3y - z = 3$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to

The point of intersection C of the plane $8x + y + 2z = 0$ and the line joining the points $A(-3, -6, 1)$ and $B(2, 4, -3)$ divides the line segment AB internally in the ratio $k:1$. If $a, b, c$ ($|a|, |b|, |c|$ are coprime) are the direction ratios of the perpendicular from the point C on the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$, then $|a+b+c|$ is equal to _____.

$P_1:\ 3x-y-7z=11$; $P_2$ through $(2,-1,0),(2,0,-1),(5,1,1)$. Foot of perpendicular from $(7,4,-1)$ on $P_1\cap P_2$ is $(\alpha,\beta,\gamma)$. Then $\alpha+\beta+\gamma$ is equal to

Plane $4x-y+z=10$ rotated by $\frac{\pi}{2}$ about its intersection with $x+y-z=4$. $\alpha$ = distance of $(2,3,-4)$ from new plane. $35\alpha$ is equal to

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

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point P. Let Q be the foot of perpendicular from the point $R(1, -1, -3)$ on the line L. If $\alpha$ is the area of triangle PQR, then $\alpha^2$ is equal to _____.

Let the equation of the plane passing through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ and parallel to the line $x + y + 2z - 7 = 0 = 2x + 3y + z - 2$ be $ax + by + cz = 65$. Then the distance of the point $(a, b, c)$ from the plane $2x + 2y - z + 16 = 0$ is ______.

Plane through intersection of $x+2y+az=2$ and $x-y+z=3$ is $5x-11y+bz=6a-1$. For $c\in\mathbb{Z}$, distance from $(a,-c,c)$ is $\frac{2}{\sqrt{a}}$. Then $a+b$ is equal to

Find the shortest distance between two skew lines given position vectors and magnitude conditions for a tetrahedron $ABCD$.

$L_1: \frac{x+3}{5}=\frac{y+1}{4}=\frac{z-2}{\alpha}$ and $L_2: 3x+2y+z-2=0=x-3y+2z-13$ coplanar. Point $P(a,b,c)$ on $L_1$ nearest to $Q(-4,-3,2)$. Then $|a|+|b|+|c|$ is equal to

Plane through $(-2,3,5)$, $\perp$ to $2x+4y+5z=8$ and $3x-2y+3z=5$: $\alpha x+\beta y+\gamma z+97=0$. Then $\alpha+\beta+\gamma$ is

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The distance of the point P(4, 6, -2) from the line passing through the point $(-3, 2, 3)$ and parallel to a line with direction ratios $3, 3, -1$ is equal to:

The line coplanar with $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ is

If the square of the shortest distance between the lines $\dfrac{x-2}{1}=\dfrac{y-1}{2}=\dfrac{z+3}{-3}$ and $\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{-5}$ is $\dfrac{m}{n}$, where $m,n$ are coprime numbers, then $m+n$ is equal to:

$N$ = foot of $\perp$ from $P(1,-2,3)$ on line through $(4,5,8)$ and $(1,-7,5)$. Distance of $N$ from $2x-2y+z+5=0$ is

If the lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1}$ and $\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is:

$a,b\in\mathbb{Z}$, $|a-b|\leq10$. Angle between plane $ax+y-z=b$ and line $x-1=a-y=z+1$ is $\cos^{-1}(\frac{1}{3})$. Distance of $(6,-6,4)$ from plane is $3\sqrt{6}$. Then $a^4+b^2$ is equal to

$\lambda_1,\lambda_2$ s.t. $(\frac{5}{2},1,\lambda)$ and $(-2,0,1)$ equidistant from $2x+3y-6z+7=0$. $\lambda_1>\lambda_2$. Distance of $(\lambda_1-\lambda_2,\lambda_2,\lambda_1)$ from line $\frac{x-1}{5}=\frac{y-2}{1}=\frac{z+7}{2}$ is ______

A tetrahedron has vertices \(P(1, 2, 1)\), \(Q(2, 1, 3)\), \(R(-1, 1, 2)\) and \(O(0, 0, 0)\). The angle between the faces \(OPQ\) and \(PQR\) 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

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

If a plane p₁ is drawn from the point A(\(\vec{a}\)) and another plane p₂ is drawn from point B(\(\vec{b}\)) parallel to p, then the distance between the planes p₁ and p₂ is:

Let the point $A$ divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $r:1$ ($r>0$). If $O$ is the origin and $\left(\overrightarrow{OQ}\cdot\overrightarrow{OA}\right)-\dfrac{1}{5}\left|\overrightarrow{OP}\times\overrightarrow{OA}\right|^2=10$, then the value of $r$ is:

Given the plane containing the line \(\dfrac{x-3}{2} = \dfrac{y+2}{-1} = \dfrac{z-1}{3}\) and also containing its projection on the plane \(2x + 3y - z = 5\). The normal vector of required plane is \((2\hat{i} - \hat{j} + 3\hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k})\). Which of the following points satisfy the required plane equation?

If the shortest distance between the lines $\dfrac{x+2}{2}=\dfrac{y+3}{3}=\dfrac{z-5}{4}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{-3}=\dfrac{z+4}{2}$ is $\dfrac{38}{3\sqrt{5}}k$, and $\displaystyle\int_0^k[x^2]\,dx=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^3$ is equal to _____

In a regular tetrahedron, let θ be the angle between any edge and a face not containing the edge. The value of \(\cos^2 θ\) is

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L:\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ and $R(5,p,q)$ be a point on $L$. Then the square of the area of $\triangle PQR$ is ________.

If N be the foot of the perpendicular from point D on the plane face ABC, then find the position vector of N.

Any plane through (1, 0, 0) is \(a(x-1) + by + cz = 0\). It passes through (0, 1, 0) and makes an angle of \(\frac{\pi}{4}\) with \(x + y = 3\). Find the ratio \(a : b : c\).

If the shortest distance between the lines \(\dfrac{x-1}{\alpha} = \dfrac{y+1}{-1} = \dfrac{z}{1}\), \((\alpha \ne -1)\) and \(x + y + z + 1 = 0 = 2x - y + z + 3\) is \(\dfrac{1}{\sqrt{3}}\), then a value of \(\alpha\) is

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\dfrac{x-3}{7}=\dfrac{y-2}{-1}=\dfrac{z+1}{-2}$. Then the area of the right angled triangle $PQR$, where $R$ is the point $(3,-2,1)$, is:

The coordinates of points, whose perpendicular distances from $yz, zx$ and $xy$-planes are in A.P., and whose distances from $x, y$ and $z$ axes are $\sqrt{13}, \sqrt{10}$ and $\sqrt{5}$ respectively is:

The straight lines whose direction cosines are given by the relations $al + bm + cn = 0$ and $fnn + gnl + hlm = 0$ are perpendicular if:

The distance of the point \((1, 3, -7)\) from the plane passing through the point \((1, -1, -1)\), having normal perpendicular to both the lines \(\dfrac{x-1}{1} = \dfrac{y+2}{-2} = \dfrac{z-4}{3}\) and \(\dfrac{x-2}{2} = \dfrac{y+1}{-1} = \dfrac{z+7}{-1}\), is

Let L be the line of intersection of the planes \(2x + 3y + z = 1\) and \(x + 3y + 2z = 2\). If L makes an angle \(\alpha\) with the positive x-axis, then \(\cos\alpha\) equals

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{b}=2\hat{i}+7\hat{j}+3\hat{k}$. Let $L_1:\overrightarrow{r}=(-\hat{i}+2\hat{j}+\hat{k})+\lambda\overrightarrow{a}$, $\lambda\in\mathbf{R}$ and $L_2:\overrightarrow{r}=(\hat{j}+\hat{k})+\mu\overrightarrow{b}$, $\mu\in\mathbf{R}$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$, and is parallel to $\overrightarrow{a}+\overrightarrow{b}$, then $L_3$ passes through the point: