3D Geometry Questions (578)

If the line $\dfrac{2-x}{3}=\dfrac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\dfrac{x+3}{3\mu}=\dfrac{1-2y}{6}=\dfrac{5-z}{7}$, then $4\lambda+9\mu$ is equal to:

Let $P(x,y,z)$ be a point in the first octant, whose projection in the $xy$-plane is the point $Q$. Let $OP=\gamma$; the angle between $OQ$ and the positive $x$-axis be $\theta$; and the angle between $OP$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is

Let the line $L$ intersect the lines $x-2=-y=z-1$, $2(x+1)=2(y-1)=z+1$ and be parallel to the line $\dfrac{x-2}{3}=\dfrac{y+1}{1}=\dfrac{z-2}{2}$. Then which of the following points lies on $L$?

One vertex of a rectangular parallelopiped is at origin $O$; edges along axes are 3, 4, 5. $P=(3,4,5)$. Shortest distance between diagonal $OP$ and an edge parallel to $z$-axis (not through $O$ or $P$) is

Plane through intersection of $2x-y+z=3$ and $4x-3y+5z+9=0$, parallel to $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$, is $ax+by+cz+6=0$. Then $a+b+c$ is

Shortest distance between $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ is

$O$ origin, $P=-\hat{i}-2\hat{j}+3\hat{k}$. $A=-2\hat{i}+\hat{j}-3\hat{k}$, $B=2\hat{i}+4\hat{j}-2\hat{k}$, $C=-4\hat{i}+2\hat{j}-\hat{k}$. Projection of $\overrightarrow{OP}$ on vector $\perp\overrightarrow{AB}$ and $\overrightarrow{AC}$ is

Line $L:\ x=\frac{1-y}{-2}=\frac{z-3}{\lambda}$ meets plane $x+2y+3z=4$ at $(\alpha,\beta,\gamma)$. Angle between $L$ and plane is $\cos^{-1}(\sqrt{\frac{5}{14}})$. Then $\alpha+2\beta+6\gamma$ is equal to

Image of $(5,5,8)$ in $x-2y+z-2=0$ is $P$. Distance of $Q(6,-2,\alpha)$, $\alpha>0$ from $P$ is $\frac{13}{3}$. Then $\alpha$ is equal to _______

$S$ = set of $\lambda$ for which SD between $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8|\sum_{\lambda\in S}\lambda|$ is equal to

Foot of $\perp$ of $P(3,-2,-9)$ on plane through $(-1,-2,-3),(9,3,4),(9,-2,1)$ is $Q(\alpha,\beta,\gamma)$. Distance of $Q$ from origin is

Plane $P$ contains $2x+y-z-3=0=5x-3y+4z+9$ and is parallel to $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Distance of $A(8,-1,-19)$ from $P$ measured $\parallel$ to $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to _________

Line $x=y=z$ intersects $x\sin A+y\sin B+z\sin C-18=0=x\sin2A+y\sin2B+z\sin2C-9$ where $A,B,C$ are angles of $\triangle ABC$. Then $80\!\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$ is equal to _________

The equation of a line is \(\dfrac{x}{1} = \dfrac{y-1}{0} = \dfrac{z+1}{-1}\). From point \(P(\beta, 0, \beta)\) where \(\beta \neq 0\), a perpendicular is drawn to the given line meeting at point \(M(\lambda, 1, -\lambda-1)\). Find the value of \(\beta\).

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

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

Let the line of the shortest distance between the lines $L_1:\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $L_2:\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha,\beta,\gamma)$ is the midpoint of the line segment $PQ$, then $2(\alpha+\beta+\gamma)$ is equal to

Let $P$ and $Q$ be the points on the line $\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}$ which are at a distance of 6 units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha,\beta,\gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is:

The perpendicular distance, of the line y+2 x-1 2 = -1 = z+3 2 from the point P(2, -10, 1) , is :

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 P QR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR) is equal to _______ -. 2

If the shortest distance between the lines $\dfrac{x-4}{1}=\dfrac{y+1}{2}=\dfrac{z}{-3}$ and $\dfrac{x-\lambda}{2}=\dfrac{y+1}{4}=\dfrac{z-2}{-5}$ is $\dfrac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is:

Let the image of the point $(1,0,7)$ in the line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$ be the point $(\alpha,\beta,\gamma)$. Then which one of the following points lies on the line passing through $(\alpha,\beta,\gamma)$ and making angles $\dfrac{2\pi}{3}$ and $\dfrac{3\pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with $x$-axis?

The position vectors of the vertices $A$, $B$ and $C$ of a triangle are $2\hat{i}-3\hat{j}+3\hat{k}$, $2\hat{i}+2\hat{j}+3\hat{k}$ and $-\hat{i}+\hat{j}+3\hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$ where $D$ is on the line segment $BC$, then $2l^2$ equals:

Let $P(3,2,3)$, $Q(4,6,2)$ and $R(7,3,2)$ be the vertices of $\triangle PQR$. Then, the angle $\angle QPR$ is

Let O be the origin, and M and N be the points on the lines $\dfrac{x-5}{4}=\dfrac{y-4}{1}=\dfrac{z-5}{3}$ and $\dfrac{x+8}{12}=\dfrac{y+2}{5}=\dfrac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overrightarrow{OM}\cdot\overrightarrow{ON}$ is equal to

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$, $x=y+2=6z-6$ and $d_2$ is the shortest distance between the lines $\dfrac{x-1}{2}=\dfrac{y+8}{-7}=\dfrac{z-4}{5}$, $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-6}{-3}$, then the value of $\dfrac{32\sqrt{3}\,d_1}{d_2}$ is:

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a,a,a)$ on the lines $x=y,z=1$ and $x=-y,z=-1$ respectively. If $\angle QPR$ is a right angle, then $12a^2$ is equal to

Let a line passing through the point $(-1,2,3)$ intersect the lines $L_1:\dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z+1}{-2}$ at $M(\alpha,\beta,\gamma)$ and $L_2:\dfrac{x+2}{-3}=\dfrac{y-2}{-2}=\dfrac{z-1}{4}$ at $N(a,b,c)$. Then the value of $\dfrac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals

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

The distance of the line x-2 y-6 z-3 y-2 z+3 2 = 3 = 4 from the point (1, 4, 0) along the line x 1 = 2 = 3 is :

Let L 1 : x-1​ 1 = y-2 y-2 -1 = 2 and L 2 : -1 = 2 z-1 ​ x+1 = z1 be two lines. ​ ​ ​ ​ ​ Let L 3 be a line passing through the point (\alpha, \beta, \gamma) and be perpendicular to both L 1 and L 2 . If L 3 ​ ​ ​ ​ intersects L 1 , then ∣5\alpha - 11\beta - 8\gamma∣ equals : ​

Let a line pass through two distinct points P (-2, -1, 3) and Q, and be parallel to the vector 3^i + 2^j + 2k ^ . If the distance of the point Q from the point R(1, 3, 3) is 5 , then the square of the area of △P QR is equal to :

Let in a △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 y-7 . Then the area (in sq. units) of △ABC is: x-6 z-7 = = 3 2 -2

If the shortest distance between the lines $\dfrac{x-\lambda}{-2}=\dfrac{y-2}{1}=\dfrac{z-1}{1}$ and $\dfrac{x-\sqrt{3}}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{1}$ is 1, then the sum of all possible values of $\lambda$ is:

Let O be the origin and the position vectors of $A$ and $B$ be $2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is

Minimum value of the expression $(a-b)^2 + \left(4\sqrt{1+a^2} + 2\sqrt{2b-b^2}\right)^2$ is

A line passes through $A(4,-6,-2)$ and $B(16,-2,4)$. The point $P(a,b,c)$ where $a,b,c$ are non-negative integers, on the line $AB$ lies at a distance of 21 units, from the point $A$. The distance between the points $P(a,b,c)$ and $Q(4,-12,3)$ is equal to

Let $L_1:\dfrac{x-1}{3}=\dfrac{y-1}{-1}=\dfrac{z+1}{0}$ and $L_2:\dfrac{x-2}{2}=\dfrac{y}{0}=\dfrac{z+4}{\alpha}$, $\alpha\in\mathbf{R}$, be two lines which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26\alpha(PB)^2$ is ________.

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

Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3\hat{i}+2\hat{j}+2\hat{k}$. If the distance of the point $Q$ from the point $R(1,3,3)$ is $5$, then the square of the area of $\triangle PQR$ is equal to:

The perpendicular distance, of the line $\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z+3}{2}$ from the point $P(2,-10,1)$, is:

Consider a line $L$ passing through the points $P(1,2,1)$ and $Q(2,1,-1)$. If the mirror image of the point $A(2,2,2)$ in the line $L$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+6\gamma$ is equal to _____

If the shortest distance between the lines $\dfrac{x-\lambda}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{1}$ and $\dfrac{x+2}{3}=\dfrac{y+5}{2}=\dfrac{z-4}{4}$ is $\dfrac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _____

Let $P(\alpha,\beta,\gamma)$ be the image of the point $Q(1,6,4)$ in the line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$. Then $2\alpha+\beta+\gamma$ is equal to _____

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

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

The coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 4z − 12 = 0 is

Ex. 44 Given the line L: \(\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-3}{-1}\) and the plane \(\pi: x - 2y - z = 0\)Statement I: L lies in \(\pi\).Statement II: L is parallel to \(\pi\).

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

Let $\vec{a}$, $\vec{b}$ be two vectors perpendicular to each other with $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{c}\times\vec{a}=\vec{b}$. The least value of $|\vec{c}-\vec{a}|$ is