3D Geometry Questions - JEE

The equation of the plane through the line of intersection of $4x + 7y + 4z + 81 = 0$ and $5x + 3y + 10z = 25$ and perpendicular to $4x + 7y + 4z + 81 = 0$ is:

82. The equation of parallel line passing through the point $(2, 3, -4)$ and parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

Given lines are $2x = 3y = -z$ and $6x = -y = -4z$. The angle between the two lines is:

Let $L$ be the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+3}{6}$ and let $S$ be the set of all points $(a,b,c)$ on $L$, whose distance from the line $\dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z-9}{0}$ along the line $L$ is 7. Then $\sum_{(a,b,c)\in S}(a+b+c)$ is equal to:

If the image of the point $P(1,2,a)$ in the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{7-z}{2}$ is $Q(5,b,c)$, then $a^2+b^2+c^2$ is equal to

If the distances of the point $(1,2,a)$ from the line $\dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$ along the lines $L_1:\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-a}{b}$ and $L_2:\dfrac{x-1}{1}=\dfrac{y-2}{4}=\dfrac{z-a}{c}$ are equal, then $a+b+c$ is equal to

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ 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

If the image of the point $(1, -2, 3)$ in the plane $2x + 3y - z = 7$ is the point $(\alpha, \beta, \gamma)$, then the value of $\alpha + \beta + \gamma$ is equal to

The foot of the point $P(1, -3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $20x = 5y - 4z$ is point $Q$, then the value of $|PQ|^2$ is

The lines \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}\ and \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}\ are

The distance between the point $(-1, -5, -10)$ and the point of intersection of the line $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-2}{-2}$ with the plane $x + y + z = 5$ is $13t$, then $t$ equals to

ABCD is a regular tetrahedron, A is the origin and B lies on x-axis. ABC lies in the xy-plane and $|\vec{AB}| = 2$. Under these conditions, the number of possible tetrahedrons is:

Let the vertices of a triangle are A$(x_1, y_1, z_1)$, B$(x_2, y_2, z_2)$ and C$(x_3, y_3, z_3)$. The mid-points of sides AB, BC and CA are F(2, 3, −1), D(5, 7, 11) and E(0, 8, 5) respectively. Find $x_1 + x_2$.

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y − z + 4 = 0 and parallel to the Y-axis also passes through the point

The vector equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $\vec{r} \cdot (\vec{i} + 2\vec{j} - 5\vec{k}) + 9 = 0$ is

The distance of a point $2, 5, -3$ from the plane $6x - 3y + 2z = 4$ is

Given two lines \[L_1: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}\]\[L_2: \frac{x-3}{1} = \frac{y-k}{2} = \frac{z-0}{1}\]Find the value of $k$ such that the lines $L_1$ and $L_2$ are intersecting each other.

Let OABC be a regular tetrahedron of edge length unity. Its volume be V and $6V = \frac{p}{q}$ where p and q are relatively prime. Then find the value of $(p + q)$:

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

Find the distance between the planes $P_1: x - 2y - 2z - 4 = 0$ and $P_2: 2x - 4y - 4z - 6 = 0$

The distance of the point having position vector $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line passing through the point $(2, 3, -4)$ and parallel to the vector, $6\hat{i} + 3\hat{j} - 4\hat{k}$ is ______.

The distance of the plane $x + 2y - z = 2$ from the point $(2, -1, 3)$, measured in the direction with the direction ratios $(2, 2, 1)$ is

The plane $3x + 4y + 12z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. If the equation of the plane in new position is $x - 4y + 6x = K$, then the value of $K$ is

The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is

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

Consider a tetrahedron D—ABC with position vectors of its angular points as A(1, 1, 1); B(1, 2, 3); C(1, 1, 2) and centre of tetrahedron \left(\frac{3}{2}, \frac{3}{4}, 2\right). Find the shortest distance between the skew lines AB and CD.

Let $P$ be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is ______ (up to three decimal places).

The position vectors of the points P and Q are $5\hat{i} + 7\hat{j} - 2\hat{k}$ and $-3\hat{i} + 3\hat{j} + 6\hat{k}$, respectively. The vector $\vec{A} = 3\hat{i} - \hat{j} + \hat{k}$ passes through the point P and the vector $\vec{B} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ passes through the point Q. A third vector $2\hat{i} + 7\hat{j} - 5\hat{k}$ intersects vectors $\vec{A}$ and $\vec{B}$. Then the distance between the two position vectors of the points of intersection is __________ (up to three decimal places).

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $P$, be $(\alpha,\beta,\gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to:

The number of planes that are equidistant from four non-coplanar points is

The shortest distance between the lines $\dfrac{x-3}{4}=\dfrac{y+7}{-11}=\dfrac{z-1}{5}$ and $\dfrac{x-5}{3}=\dfrac{y-9}{-6}=\dfrac{z+2}{1}$ is:

A line makes an angle $\theta$ both with X and Y-axes. A possible value of $\theta$ is in

The shortest distance between the line $x = y = z$ and the line of intersection of $2x + y + z - 1 = 0$ and $3x + y + 2z - 2 = 0$ is

Find the value of $\lambda$ if the plane $2x - 2y + z - 3 + \lambda = 0$ contains the point $(3, 1, 1)$.

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

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

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

The distance of the point $(-1, 9, -16)$ from the plane $2x + 3y - z = 5$ measured parallel to the line $\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$ is:

Consider the lines $L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$ and $L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$. A line $L_3$ having direction ratios $1, -1, -2$, intersects $L_1$ and $L_2$ at the points P and Q respectively. Then the length of line segment PQ is

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

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r}\cdot(3\hat{i}-5\hat{j}+\hat{k}) = 7$ and $P_2: \vec{r}\cdot(\lambda\hat{i}+\hat{j}-3\hat{k}) = 9$ is $\sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)$, then the square of the length of perpendicular from the point $(38\lambda_1, 10\lambda_2, 2)$ to the plane $P_1$ is _____.

Let a line L pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of L from the point $(5, 3, 8)$ is $\alpha$, then $3\alpha^2$ is equal to _____.

Let the shortest distance between the lines $L: \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$, $\lambda \geq 0$ and $L_1: x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on L, then which of the following is NOT possible?

Let the image of the point P(2, -1, 3) in the plane $x + 2y - z = 0$ be Q. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point Q is

Let $\alpha x + \beta y + yz = 1$ be the equation of a plane passing through the point $(3, -2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 5)$. Then the value of $\alpha\beta y$ is equal to _____.

Let $(\alpha,\beta,\gamma)$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}$. Then $19(\alpha+\beta+\gamma)$ is equal 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

If the length of the perpendicular from point $P(a, 4, 2)$, $a>0$, to the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}$ is $2\sqrt{6}$ units, and $Q(\alpha_1,\alpha_2,\alpha_3)$ is the image of $P$ on this line, then $a+\displaystyle\sum_{i=1}^3 \alpha_i$ equals

Let $P$ be the foot of the perpendicular from the point $M(1,2,2)$ on the line $L:\dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z-2}{2}$. Let the line $\overrightarrow{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$, $\lambda\in\mathbf{R}$, intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to:

3D Geometry Questions (578)