3D Geometry Questions (578)

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

Assuming the plane \(4x - 3y + 7z = 0\) to be horizontal, the direction cosines of the line of greatest slope in the plane \(2x + y - 5z = 0\) are

Consider the line $L$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\dfrac{11}{3},\dfrac{11}{3},\dfrac{19}{3}\right)$ from the line $L$ along the line $\dfrac{3x-11}{2}=\dfrac{3y-11}{1}=\dfrac{3z-19}{2}$ is equal to:

The minimum value of $x^2+y^2+z^2$ if $ax+by+cz=p$ 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

The shortest distance between the lines \(\frac{x - 3}{7} = \frac{y - 8}{-6} = \frac{z - 3}{1}\) and \(\frac{x + 3}{1} = \frac{y + 7}{-2} = \frac{z - 6}{1}\) is

The length of the perpendicular from the point \((2, -1, 4)\) on the straight line \(\frac{x + 3}{10} = \frac{y - 2}{-7} = \frac{z}{1}\) is

A plane which passes through the point (3, 2, 0) and the line \(\dfrac{x-4}{1} = \dfrac{y-7}{5} = \dfrac{z-4}{4}\) 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

The equation of the plane containing line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to plane $4x+5y-3z-8=0$ is

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

Consider the set of eight vectors $V=\{a\hat{i}+b\hat{j}+c\hat{k}: a,b,c\in\{-1,1\}\}$. The number of ways three non-coplanar vectors can be chosen from $V$ equals

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

A ladder of 3m length leans against a wall. The ladder forms a vertical angle of 30° with the wall. The top slides down at 20 cm/s. The bottom slides away at 20 cm/s at time $t$. The average velocity of a person halfway up the ladder for the first $t$ seconds is

Let $Q$ be the cube with vertices $\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1,x_2,x_3\in\{0,1\}\}$. Let $F$ be the set of all 12 lines containing face diagonals and $S$ be the set of 4 main diagonals. For lines $l_1\in F$ and $l_2\in S$, let $d(l_1,l_2)$ denote shortest distance. Maximum of $d(l_1,l_2)$ is $\lambda$. Find $\lambda^{-2}$.

Let \(A_1, A_2, A_3, A_4\) be the areas of the triangular faces of a tetrahedron, and \(h_1, h_2, h_3, h_4\) be the corresponding altitude of the tetrahedron. If volume of tetrahedron is \(1/6\) cubic units, then find the minimum value of \((A_1 + A_2 + A_3 + A_4)(h_1 + h_2 + h_3 + h_4)\) (in cubic units).

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

The line $(x,y,z)=(2,-3,4)+\lambda(1,2,-3)$ intersects $2x+3y-z=13$ at $P$ and $yz$-plane at $Q$. If $PQ=a\sqrt{b}$, $a,b\in\mathbb{N}$, $a>3$, then $\dfrac{a+b}{3}$ equals

The equation of the plane passing through $(1,-1,2)$ and perpendicular to planes $2x+3y-2z=5$ and $x+2y-3z=8$ is

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

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

Statement-1: The point \(A(3, 1, 6)\) is the mirror image of the point \(B(1, 3, 4)\) in the plane \(x - y + z = 5\).Statement-2: The plane \(x - y + z = 5\) bisects the line segment joining \(A(3, 1, 6)\) and \(B(1, 3, 4)\).

A plane meets coordinate axes at $A$, $B$, $C$ such that centroid of $\triangle ABC$ is at $(p,q,r)$. Equation of plane is

The angle between two lines whose direction cosines satisfy $l+m+n=0$ and $l^2=m^2+n^2$ is

If $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+7\hat{j}+2\hat{k}$, $\vec{x}\cdot\vec{a}=0$ and $\vec{x}\cdot\vec{c}=0$ for some non-zero vector $\vec{x}$, then value of $\vec{a}\cdot(\vec{b}\times\vec{c})$ is

The distance of point $P(3,8,2)$ from the line $\dfrac{x-1}{2}=\dfrac{y-3}{4}=\dfrac{z-2}{3}$ measured parallel to plane $3x+2y-2z+15=0$ is

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

Let $\vec{p},\vec{q}$ and $\vec{r}$ be three unit vectors satisfying $|\vec{p}-\vec{q}|^2+|\vec{q}-\vec{r}|^2+|\vec{r}-\vec{p}|^2=9$. Then $|2\vec{p}+5\vec{q}+5\vec{r}|$ is equal to

Match List-I with List-II: (A) Line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-2k}{2}$ lies in plane $2x-4y+z=3$; (B) Lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, value of $k$; (C) Plane through $(1,1,1)$ with OA=OB=OC, volume of tetrahedron OABC; (D) Distance from $(-1,5/\sqrt{2},3/\sqrt{2})$ to plane $P$ through $(1,-2,1)$ perpendicular to $2x-2y+z=0$ and $x-y+2z=4$