3D Geometry Questions (578)

$L_1: \frac{x+3}{5}=\frac{y+1}{4}=\frac{z-2}{\alpha}$ 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

3825

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:

Let the equation of the plane be \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\). The volume of tetrahedron \(OABC\) is \(V = \frac{1}{6}(abc)\). Given that \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\), find the minimum value of \(V\).

The square of the distance of the point $\left(\dfrac{15}{7},\dfrac{32}{7},7\right)$ from the line $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ in the direction of the vector $\hat{i}+4\hat{j}+7\hat{k}$ is:

Let PM be perpendicular from the point $P(1, 2, 3)$ to $x$-$y$ plane. If $\overrightarrow{OP}$ makes an angle $\theta$ with the positive direction of $z$-axis and $\overrightarrow{OM}$ makes an angle $\phi$ with the positive direction of $x$-axis, where $O$ is the origin and $\theta$ and $\phi$ are acute angles, then:

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 $PQR$ is formed such that $Q$ lies on one of the parallel lines, while $R$ lies on the other. Then $(QR)^2$ is equal to ________.

If the angle \(\theta\) between the line \(\dfrac{x+1}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{2}\) and the plane \(2x - y + \sqrt{\lambda}z + 4 = 0\) is such that \(\sin\theta = \dfrac{1}{3}\), the value of \(\lambda\) is

The number of distinct real values of \(\lambda\) for which the lines \(\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z+3}{\lambda^2}\) and \(\dfrac{x-3}{1} = \dfrac{y-2}{\lambda^2} = \dfrac{z-1}{2}\) are coplanar is

Let $P$ be the point of intersection of the lines $\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{1}$ and $\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{2}$. Then the shortest distance of $P$ from the line $4x=2y=z$ is:

Let $P$ be the point $(10,-2,-1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1,7,6)$ on the line passing through the points $(2,-5,11)$ and $(-6,7,-5)$. Then the length of the line segment $PQ$ is equal to _____

Angle between the line \(\dfrac{x+1}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{2}\) and plane \(2x - y + \sqrt{\lambda}z + 4 = 0\) is given by \(\sin\theta = \dfrac{1}{3}\). Find \(\lambda = ?\)

If the shortest distance between the lines $\dfrac{x-\lambda}{2}=\dfrac{y-4}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-2}{4}=\dfrac{y-4}{6}=\dfrac{z-7}{8}$ is $\dfrac{13}{\sqrt{29}}$, then a value of $\lambda$ is:

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ intersect the line $\dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is:

The vertices of △ABC are A(2, 0, 0), B(0, 1, 0), C(0, 0, 2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A, B, C and the origin O. The z-coordinate of H is:

Let in a $\triangle ABC$, the length of the side $AC$ be 6, the vertex $B$ be $(1,2,3)$ and the vertices $A$, $C$ lie on the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is:

A plane which bisects the angle between the two given planes \(2x - y + 2z - 4 = 0\) and \(x + 2y + 2z - 2 = 0\), passes through the point:

If \(A(3, 2, 0)\), \(B(5, 3, 2)\) and \(C(-9, 6, -3)\) are three points forming a triangle and \(AD\) is the bisector of \(\angle BAC\), then coordinates of \(D\) are

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 \frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3} and \frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1} is(JEE Main 2017)

Let A(x, y, z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and ( 0, 0, 1 ). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements (S1) : △ABC is an isosceles right angled triangle, and 9\sqrt2 (S2) : the area of △ABC is 2 ,

Find the point where the line $$\frac{x - 2}{3} = \frac{y - 1}{4} = \frac{z - 6}{5}$$ intersects the plane x + y - 2z = 3.

Let the point $(-1,\alpha,\beta)$ lie on the line of the shortest distance between the lines $\dfrac{x+2}{-3}=\dfrac{y-2}{4}=\dfrac{z-5}{2}$ and $\dfrac{x+2}{-1}=\dfrac{y+6}{2}=\dfrac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to _____

The centre of a sphere is \(C = (-2, 1, +3)\). The distance of centre \(C\) from plane \(12x + 4y + 3z = 327\) and the radius of the sphere is \(\sqrt{4 + 1 + 4 + 155} = 13\). The shortest distance from the plane to the sphere is:

On which of the following lines lies the point of intersection of the line, \(\dfrac{x-4}{2} = \dfrac{y-5}{2} = \dfrac{z-3}{1}\) and the plane, \(x + y + z = 2\)?

The length of perpendicular from P(2, -3, 1) to the line \(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1}\) is