3D Geometry Questions - JEE

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

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

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

If direction cosines of line $L$ be $l, m, n$, and $2l + 3m + n = 0$ and $l + 3m + 2n = 0$, then what is $\cos\alpha$ where $\alpha$ is the angle the line makes with the x-axis?

For lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{\lambda^2}$ and $\frac{x-3}{1} = \frac{y-2}{\lambda^2} = \frac{z-1}{2}$ to be coplanar, the number of values of $\lambda$ is:

The vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ is

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ have magnitudes 1, 2, 3 respectively, satisfying $|(\vec{a}\times\vec{b})\cdot\vec{c}|=6$. If $\vec{d}$ is a unit vector coplanar with $\vec{b}$ and $\vec{c}$ such that $\vec{b}\cdot\vec{d}=1$, then $|(\vec{a}\times\vec{c})\cdot\vec{d}|^2+|(\vec{a}\times\vec{c})\times\vec{d}|^2$ is

In a cubical hall ABCD-PQRS with each side 10 m, G is the centre of the wall BCRQ and T is the mid-point of the side AB. The angle of elevation of G at the point T is

The centre of the sphere (x-4)(x+4) + (y-3)(y+3) + z^2 = 0 is

The equation of the plane $P_1$ parallel to line $L: \frac{x-1}{2}=\frac{y-0}{3}=\frac{z-1}{1}$, at distance 6 from the intersection of $L$ with plane $P: 2x+3y+z-18=0$, and having maximum distance from $Q(5,4,-4)$, is $ax+by+cz+d=0$. The value of $\dfrac{bd}{ac}$ is:

Let $\gamma\in\mathbb{R}$ such that $L_1: \frac{x+1}{11}=\frac{y+2}{21}=\frac{z+3}{29}$ and $L_2: \frac{x+3}{16}=\frac{y+2}{11}=\frac{z+4}{\gamma}$ intersect at $R_1$. Match P)$\gamma$; Q)unit normal $\hat{n}$; R)$\overrightarrow{OR_1}$; S)$\overrightarrow{OR_1}\cdot\hat{n}$ with List-II: 1)$-\hat{i}-\hat{j}+\hat{k}$; 2)$\frac{3}{2}$; 3)$1$; 4)$\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$; 5)$\frac{2}{3}$

There is a plane $P$ passing through the line $\dfrac{x+1}{2}=\dfrac{y-1}{1}=\dfrac{z+3}{2}$ and parallel to the $x$-axis. Let $\vec{v}$ be a vector parallel to plane $P$ such that the angle between $\vec{v}$ and $\vec{u}=\hat{i}+\hat{j}+\hat{k}$ is least. There is a circle lying on plane $P$ with centre $C(0,0,-5)$ and radius $\sqrt{10}$. A line $L$ on plane $P$ is tangent to the circle at a point $Q(x,y,z)$ with the least possible value of $z$. Then which of the following is/are true?

Let $\vec{a},\vec{b},\vec{c},\vec{d}$ be position vectors of $A,B,C,D$ with $|\vec{a}|=1$, $|\vec{b}|=2$, $|\vec{c}|=3$, $|\vec{d}|=4$, $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=24$, $\vec{a}\cdot\vec{c}=\frac{3}{2}$, $\vec{b}\cdot\vec{c}>0$. Area of $\triangle ABC=\frac{1}{4}(2\alpha+\beta\sqrt{3})$ and area of $\triangle BCD=\frac{1}{2}(m-n\sqrt{3})$ where $\alpha,\beta,m,n\in\mathbb{N}$. Match P)$\alpha$, Q)$\beta$, R)$m$, S)$n$ with 1)1, 2)3, 3)4, 4)9.

There is a plane $P$ passing through the line $\dfrac{x+1}{2}=\dfrac{y-1}{1}=\dfrac{z+3}{2}$ and parallel to the $x$-axis. Let $\vec{v}$ be a vector parallel to plane $P$ such that the angle between $\vec{v}$ and $\vec{u}=\hat{i}+\hat{j}+\hat{k}$ is least. There is a circle lying on plane $P$ with centre $C(0,0,-5)$ and radius $\sqrt{10}$. A line $L$ on plane $P$ is tangent to the circle at a point $Q(x,y,z)$ with the least possible value of $z$. Then which of the following is/are true?

Let $\vec{a},\vec{b},\vec{c},\vec{d}$ be position vectors of $A,B,C,D$ with $|\vec{a}|=1$, $|\vec{b}|=2$, $|\vec{c}|=3$, $|\vec{d}|=4$, $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=24$, $\vec{a}\cdot\vec{c}=\frac{3}{2}$, $\vec{b}\cdot\vec{c}>0$. Area of $\triangle ABC=\frac{1}{4}(2\alpha+\beta\sqrt{3})$ and area of $\triangle BCD=\frac{1}{2}(m-n\sqrt{3})$ where $\alpha,\beta,m,n\in\mathbb{N}$. Match P)$\alpha$, Q)$\beta$, R)$m$, S)$n$ with 1)1, 2)3, 3)4, 4)9.

Let $\gamma\in\mathbb{R}$ such that $L_1: \frac{x+1}{11}=\frac{y+2}{21}=\frac{z+3}{29}$ and $L_2: \frac{x+3}{16}=\frac{y+2}{11}=\frac{z+4}{\gamma}$ intersect at $R_1$. Match P)$\gamma$; Q)unit normal $\hat{n}$; R)$\overrightarrow{OR_1}$; S)$\overrightarrow{OR_1}\cdot\hat{n}$ with List-II: 1)$-\hat{i}-\hat{j}+\hat{k}$; 2)$\frac{3}{2}$; 3)$1$; 4)$\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$; 5)$\frac{2}{3}$

Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3), D(x4, y4, z4) be the vertices of a tetrahedron. If E is the centroid of face BCD and G is the centroid of ABCD, then find the value of K such that AG = K(AE).

The equation of plane containing line AC and at a maximum distance from B isGiven: A = 3i + 4j, C = 4i + 3j, B = 7i + 7j

The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $\pi/4$ with the plane $y - z + 5 = 0$ are:

If a line makes angles α, β and γ with the coordinate axes, then:

The equation of the straight line through the origin parallel to the line $(b+c)x + (c+a)y + (a+b)z = k = (b-c)x + (c-a)y + (a-b)z$ is:

The locus of intersection of locus of $P$ with $2x + y + z = 2$ is:

If the line $\frac{x-2}{-1} = \frac{y+2}{1} = \frac{z+k^2-1}{4}$ is one of the angle bisector of the lines $\frac{x}{1} = \frac{y}{-2} = \frac{z}{3}$ and $\frac{x}{-2} = \frac{y}{3} = \frac{z}{1}$, then the value of $k$ is/are:

Plane $P$ through $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and $(2,4,-3)$. Image of $(-1,3,4)$ in $P$ is $(\alpha,\beta,\gamma)$. Then $\alpha+\beta+\gamma$ is equal to

The equations of the planes through the origin which are parallel to the line $\frac{x - 1}{2} = \frac{y + 3}{-1} = \frac{z + 1}{-2}$ and at a distance $5/3$ from it are:

A tetrahedron has vertices at $O(0, 0, 0)$, $A(1, 2, 1)$, $B(2, 1, 3)$ and $C(-1, 1, 2)$. Then, the angle between the faces $OAB$ and $ABC$ will be

A line makes the same angle θ with X-axis and Z-axis. If the angle β, which it makes with Y-axis, is such that $\sin 2β = 3 \sin 2θ$, then the value of $\cos^2 θ$ is

Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $\dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$. Then the distance of $Q$ from the line $\dfrac{x-9}{3}=\dfrac{y-9}{1}=\dfrac{z-5}{-2}$ is

3D Geometry Questions (578)