3D Geometry Questions (578)

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
If \(L_1\) is the line of intersection of the planes \(2x - 2y + 3z - 2 = 0\), \(x - y + z + 1 = 0\) and \(L_2\) is the line of intersection of the planes \(x + 2y - z - 3 = 0\), \(3x - y + 2z - 1 = 0\), then the distance of the origin from the plane, containing the lines \(L_1\) and \(L_2\) 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
If the line \(\dfrac{x-3}{2} = \dfrac{y+2}{-1} = \dfrac{z+4}{3}\) lies in the plane \(lx + my - z = 9\), then \(l^2 + m^2\) is equal to
Let the equation of plane P be (2x + 3y + z + 5) + λ(x + y + z - 6) = 0. Given that the plane is perpendicular to the xy-plane, find the distance of point (0, 0, 256) from plane P (rounded to 3 decimal places).
Volume of parallelopiped determined by vectors $\vec{a},\vec{b},\vec{c}$ is 5. Then volume determined by $3(\vec{a}+\vec{b})$, $(\vec{b}+\vec{c})$ and $2(\vec{c}+\vec{a})$ is
The two lines x = ay + b, z = cy + d and x = a'y + b', z = c'y + d' will be perpendicular, if and only if
Given lines: \(x = ay + b,\; z = cy + d\) and \(x = a'y + b',\; z = c'y + d'\). These lines will be perpendicular to each other if:
A perpendicular is drawn from a point on the line \(\dfrac{x-1}{2} = \dfrac{y+1}{-1} = \dfrac{z}{1}\) 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 co-ordinates of \(Q\) are:
The given line is \(\frac{x-3}{1} = \frac{y+2}{-1} = \frac{z+\lambda}{-2} = k\). If a point P on the line lies on the plane \(2x - 4y + 3z = 2\), find \(\lambda\) and then find the shortest distance \(d\) between the two given lines. What is \(d^2\)?
The distance between the parallel planes \(2x + y + 2z = 8\) and \(2x + y + 2z = -\frac{5}{2}\) is:
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
Two lines \(\dfrac{x-3}{1} = \dfrac{y+1}{3} = \dfrac{z-6}{-1}\) and \(\dfrac{x+5}{7} = \dfrac{y-2}{-6} = \dfrac{z-3}{4}\) intersect at the point R. The reflection of R in the \(xy\)-plane has coordinates:
If a line makes an angle of \(\pi/4\) with the positive directions of each of x-axis and y-axis, then the angle that the line makes with the positive direction of the z-axis is
Distance between two parallel planes \(2x + y + 2z = 8\) and \(4x + 2y + 4z + 5 = 0\) is
The equations of lines are: \(x - ay - b = 0,\; cy - z + d = 0\) and \(x - a'y - b' = 0,\; c'y - z + d' = 0\). If these two lines are perpendicular, then which of the following is correct?
An angle between the plane, \(x + y + z = 5\) and the line of intersection of the planes, \(3x + 4y + z - 1 = 0\) and \(5x + 8y + 2z + 14 = 0\), is
The given line is \(x = 4y+5,\; z = 3y-6\). A point on the line is \((4\lambda+5,\;\lambda,\;3\lambda-6)\). The distance between the point \((4\lambda+5,\;\lambda,\;3\lambda-6)\) and \((5,3,-6)\) is 3 units. Find the point on the line closest to \((5,3,-6)\).
The line \(\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-2}{4}\) meets the plane \(x + 2y + 3z = 15\) at a point P. The distance of P from the origin is:
If the equation of the plane passing through the point \((-1, 2, 0)\) and parallel to the lines \(\dfrac{x}{3} = \dfrac{y+1}{0} = \dfrac{z-2}{-1}\) and \(\dfrac{x-1}{1} = \dfrac{y+1}{2} = \dfrac{z+1}{-1}\) is \(ax + by + cz = 1\), then the value of \((a + b + c)\), is:
If the equation of the plane through $(-1,2,0)$ and parallel to lines $\dfrac{x}{3}=\dfrac{y+1}{0}=\dfrac{z-2}{-1}$ and $\dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z+1}{-1}$ is $ax+by+cz=1$, then $a+b+c$ is
A plane bisects the line segment joining the points \((1, 2, 3)\) and \((-3, 4, 5)\) at right angles. Then this plane also passes through the point
Let the line \(\dfrac{x-2}{3} = \dfrac{y-1}{-5} = \dfrac{z+2}{2}\) lie in the plane \(x + 3y - \alpha z + \beta = 0\). Then \((\alpha, \beta)\) equals