3D Geometry Questions (578)

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

The equation of the plane which is equidistant from lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{1} = \frac{z-1}{-2}$ is $Ax + By + Cz + D = 0$ then $|A + B + C + D| = $ __________.

An angle between the lines whose direction cosines are given by the equations, \(l + 3m + 5n = 0\) and \(5lm - 2mn + 6nl = 0\), is

If the angle between the line \(x - \dfrac{y-1}{-2} - \dfrac{z-3}{\lambda}\) and the plane \(x + 2y + 3z = 4\) is \(\cos^{-1}\!\left(\sqrt{5/14}\right)\), then \(\lambda\) equals

Let $A$ be a point $(1, 2, 3)$ in the given reference system. Then locus of the point $P$ in the first octant satisfying the equation $d(O, P) = d(A, P)$ does not contain:

A variable plane makes intercepts on the co-ordinate axes the sum of whose squares is constant and equal to $k^2$. Then the locus of the foot of the perpendicular from the origin to the plane is

The equation of a line on the plane $x + y + z = 1$ such that the line $\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-1}{1}$ and the required line form a plane which is perpendicular to the plane $x + y + z = 1$ is:

The sum of all values of $\alpha$, for which the shortest distance between the lines $\dfrac{x+1}{\alpha}=\dfrac{y-2}{-1}=\dfrac{z-4}{-\alpha}$ and $\dfrac{z}{\alpha}=\dfrac{y-1}{2}=\dfrac{z-1}{2\alpha}$ is $\sqrt{2}$, is

The measure made by the $x, y$ and $z$ axis, by the plane which bisects the line joining the points $(1, 2, 3)$ and $(-3, 4, 5)$ at right angles, are $\alpha$, $\beta$ and $\gamma$ respectively, then the ordered triplet $(\alpha, \beta, \gamma)$ is

The plane $2x - y + z = 4$ intersects the line segment joining the points $A(a, -2, 4)$ and $B(2, b, -3)$ at the point C in the ratio $2:1$ and the distance of the point C from the origin is $\sqrt{5}$. If $ab < 0$ and P is the point $(a-b, b, 2b-a)$ then $CP^2$ is equal to:

If \(l + 3m + 5n = 0\) and \(5lm - 2mn + 6nl = 0\), then the directions of the two lines satisfying these conditions give \(m =\):

The line \(\dfrac{x-3}{2} = \dfrac{y+2}{-1} = \dfrac{z+4}{3}\) lies in the plane \(lx + my - z = 9\). Find \(l^2 + m^2\).

Shortest distance between the lines $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{1}$ is equal to:

Let the direction cosines of two lines satisfy the equations: $4l+m-n=0$ and $2mn+10nl+3lm=0$. Then the cosine of the acute angle between these lines is:

Let $(\alpha,\beta,\gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})$. Then the length of the projection of the vector $\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ on the vector $6\hat{i}+2\hat{j}+3\hat{k}$ is:

If coordinates of points A, B, C and D are (1, 2, 3); (4, 5, 7); (−4, 3, −6) and (2, 9, 2), respectively, then the angle between AB and CD is __________.

The vertices \(B\) and \(C\) of a \(\triangle ABC\) lie on the line, \(\dfrac{x+2}{3} = \dfrac{y-1}{0} = \dfrac{z}{4}\) such that \(BC = 5\) units. Then the area (in sq. units) of this triangle, given that the point \(A(1, -1, 2)\), is ______ (up to three decimal places).

Equation of plane containing the line $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ at minimum possible distance from $(-1,-3,1)$ is

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

A line through $(2,-2,5)$ with direction $(1,-3,2)$ meets the plane $2x-3y+4z=163$ at $P$ and the $YZ$-plane at $Q$. If $PQ = a\sqrt{b}$ where $a>3$, find $a+b$.

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 point $A$ lies on $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and point $B$ lies on $\dfrac{x-2}{3}=\dfrac{y-4}{7}=\dfrac{z-6}{6}$, then $\overrightarrow{AB}$ cannot be parallel 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