3D Geometry Questions (578)

The equation of planes bisecting the angle between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$ is/are:

Let $A$ be the point $(1, 2, 3)$ and $B$ be a point on the line $\frac{x-1}{2} = \frac{y+1}{2} = \frac{z-3}{2} = k$. The value of $k$ such that line $AB$ is perpendicular to the plane $4x + 9y - 18z = 6$ is

The value of $\sin^{-1} \sin \lambda$ is equal to:

Let P be the plane, passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of P from the point $(3, -2, 2)$ is

Projection of line $\frac{x+1}{2} = \frac{y+1}{-1} = \frac{z+3}{4}$ on the plane $x + 2y + z = 6$ has equation:

If a point $P(\alpha, \beta, \gamma)$ satisfying $(\alpha\ \beta\ \gamma)\begin{pmatrix}2&10&8\\9&3&8\\8&4&8\end{pmatrix} = (0\ 0\ 0)$ lies on the plane $2x + 4y + 3z = 5$, then $6\alpha + 9\beta + 7\gamma$ is equal to:

The vector equation of the plane through the point \(\vec{i} + 2\vec{j} - \vec{k}\) and perpendicular to the line of intersection of the planes \(\vec{r} \cdot (3\vec{i} - \vec{j} + \vec{k}) = 1\) and \(\vec{r} \cdot (\vec{i} + 4\vec{j} - 2\vec{k}) = 2\), is

A perpendicular is drawn from a point on the line $\frac{x-1}{2} = \frac{y-1}{-1} = \frac{z}{-1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x + y - z = 3$. Then the coordinates of $Q$ are

If (a, b, c) is the image of the point (1, 2, -3) in the line \(\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}\), then a + b + c is equal to

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 line passing through the points (5, 1, a) and (3, b, 1) crosses the yz-plane at the point \(\left(0, \dfrac{17}{2}, \dfrac{-13}{2}\right)\). Then

The locus of intersection of locus of $P$ and the plane $x + y + z = 2$

The equation of line intersecting and perpendicular to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and passing through $(-2, -5, 7)$ is :

The equation of the plane containing the line \(\frac{x-4}{1} = \frac{y-7}{5} = \frac{z-4}{4}\) and passing through the point (3, 2, 0) is:

If a variable point $P$ moves such that the line passing through $P$ and $Q(0, 0, 2)$ makes an angle $60°$ with $z$-axis, then locus of $P$ is:

The equation of the plane containing the line \(\dfrac{x - x_1}{l} = \dfrac{y - y_1}{m} = \dfrac{z - z_1}{n}\) is \(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\), where

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

Line \(PQ: \dfrac{x+1}{4} = \dfrac{y-2}{-2} = \dfrac{z+3}{6}\). A point \(R\) satisfies \(2x - 2y + 3z + 1 = 0\). For \(|PQ - QR|\) to be maximum, \(P\), \(Q\), \(R\) must be collinear. Find \(x_0 + y_0 + z_0\).

A perpendicular is drawn from a point $(1, 6, 3)$ to the line $\frac{x}{1} = \frac{y-2}{2} = \frac{z-3}{-3}$. What will be coordinates of the foot of perpendicular:

The foot of perpendicular of the point $(2, 0, 5)$ on the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $(\alpha, \beta, \gamma)$. Then which of the following is NOT correct?

A ray of light is coming along the line $\frac{x-2}{3} = \frac{y-1}{4} = \frac{z-6}{5}$ and strikes the plane mirror kept along the plane through the points $(2, 1, 0)$, $(5, 0, 1)$ and $(4, 1, 1)$. Then the equation of reflected ray is:

The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$, $(-1, 1, -2)$ and $(3, -4, 2)$ is:

The point of intersection of the line, passing through $(0, 0, 1)$ and intersecting the lines $x + 2y + z = 1, -x + y - 2z = 2$ and $x + y = 2, x + z = 2$ with plane is:

An equilateral triangle has its vertices on the axes of coordinates and area $\sqrt{3}$ square units. The coordinates of the orthocenter of the triangle are:

The direction cosines of the projection of the line $\frac{1}{2}(x-1) = -y = z+2$ on the plane $2x + y - 3z = 4$ are :

The plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y - z + 4 = 0\) and parallel to \(y\)-axis also passes through the point:

The equations of the lines of shortest distance between the lines $\frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$ and $\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ are:

Locus of point $P$ if $d(O, P) = k$, where $k$ is a positive constant number, represents:

If a plane passes through the points $(-1, k, 0)$, $(2, k, -1)$, $(1, 1, 2)$ and is parallel to the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is

A variable plane passes through a fixed point $(a, b, c)$ and meets the coordinate axes in $A, B, C$. The locus of the point common to the plane and also planes through $A, B, C$ parallel to coordinate planes is:

The line $\frac{x + 6}{5} = \frac{y + 10}{3} = \frac{z + 14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7, 2, 4)$. The equations of the remaining sides are:

The shortest distance between any two opposite edges of the tetrahedron is:

The plane $x = 0$ is rotated through an angle $\alpha$ about its line of intersection with the plane $z = 0$. Then equation of the plane in new position is (are):

Let $P(\alpha,\beta,\gamma)$ be the point on the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=z$ at a distance $4\sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance between the lines $\dfrac{x-\alpha}{1}=\dfrac{y-\beta}{2}=\dfrac{z-\gamma}{3}$ and $\dfrac{x+5}{2}=\dfrac{y-10}{1}=\dfrac{z-3}{1}$, is equal to

The plane $3y + 4z = 0$ is rotated about its line of intersection with the plane $x = 0$ through an angle $60°$. The equation of the plane in its new position is:

Line through $P(2,-1,2)$ and $Q(5,3,4)$ meets $x-y+z=4$ at $R$. Distance of $R$ from $x+2y+3z+2=0$ measured $\parallel$ to $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ is

Let the plane containing the line of intersection of the planes P1: $x + (\lambda+4)y + z = 1$ and P2: $2x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane P2 is

A ray $M$ is sent along the line $\frac{x - 0}{2} = \frac{y - 2}{2} = \frac{z - 1}{0}$ and is reflected by the plane $x = 0$ at point $A$. The reflected ray is again reflected by the plane $x + 2y = 0$ at point $B$. The initial ray and final reflected ray meets at point $J$. Then:

Direction cosines of normal to the plane containing lines $x = y = z$ and $x - 1 = y - 1 = \frac{z-1}{d}$ (where $d \in \mathbb{R} - \{1\}$), are :

In a three dimensional co-ordinate system $P, Q$ and $R$ are images of a point $A(a, b, c)$ in the $x-y$, the $y-z$ and the $z-x$ planes respectively. If $G$ is the centroid of triangle $PQR$ then area of triangle $AOG$ is : ($O$ is the origin)

The equation of the plane bisecting the acute angle between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$ is :

Consider the planes $P_1: 2x - y + z = 6$ and $P_2: x + 2y - z = 4$ having normals $\vec{N_1}$ and $\vec{N_2}$ respectively. The distance of the origin from the plane passing through the point $(1, 1, 1)$ and whose normal is perpendicular to $\vec{N_1}$ and $\vec{N_2}$ is

The direction cosines $l$, $m$ and $n$ of two lines are connected by the relations $l + m + n = 0$ and $lm = 0$, then the angle between the lines is

The image of the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ in the plane $x + 2y + z = 12$ is:

Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2,r,1)$ is $\sqrt{\dfrac{14}{3}}$, then the sum of all possible values of $r$ is:

Consider the lines $\frac{x}{2} = \frac{y}{3} = \frac{z}{5}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ the equation of the line which:

The distance of the point $(1, 2, 3)$ from the plane $x + y - z = 5$ measured along the straight line $z = y = 2x$ is

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three non-zero vectors satisfying $\vec{a}=\vec{b}\times\vec{c}+2\vec{b}$, where $|\vec{b}|=|\vec{c}|=2$ and $|\vec{a}|\leq4$. The sum of possible values of $|2\vec{a}+\vec{b}+\vec{c}|$ is

If $l, m, n$ represent direction cosines (if we can call it) of a vector $\overrightarrow{OP}$, then which of the following relations holds?

The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is