3D Geometry Questions (578)

Distance of $(-1,2,3)$ from $\vec{r}\cdot(\hat{i}-2\hat{j}+3\hat{k})=10$ measured $\parallel$ to SD line between $\vec{r}=(\hat{i}-\hat{j})+\lambda(2\hat{i}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

Let the equation of the plane P containing the line $x + 10 = \frac{8-y}{2} = z$ be $ax + by + 3z = 2(a+b)$ and the distance of the plane P from the point $(1, 27, 7)$ be $c$. Then $a^2 + b^2 + c^2$ is equal to _____.

Image of $P(2,3,5)$ in $2x+y-3z=6$ is $(\alpha,\beta,\gamma)$. Then $\alpha+\beta+\gamma$ is equal to

A plane passes through $(1, -2, 1)$ and is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the plane from the origin $(0, 2)$ is

Let $\theta$ be the angle between the planes $P_1 = \vec{r}\cdot(\hat{i}+\hat{j}+2\hat{k})=9$ and $P_2 = \vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k})=15$. Let L be the line that meets $P_2$ at the point $(4, -2, 5)$ and makes an angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between L and $P_2$ then $(\tan^2\theta)(\cot^2\alpha)$ is equal to _____.

Two vertices of $\triangle ABC$: $(2,4,6)$, $(0,-2,-5)$; centroid $(2,1,-1)$. Third vertex $C=(4,1,2)$. Image of $C$ in $x+2y+4z=11$ is $(\alpha,\beta,\gamma)$. $\alpha\beta+\beta\gamma+\gamma\alpha$ is equal to

Let $A(x,y,z)$ be a point in $xy$-plane, which is equidistant from three points $P(0,3,2)$, $Q(2,0,3)$ and $R(0,0,1)$. Let $B=(1,4,-1)$ and $C=(2,0,-2)$. Then among the statements (S1): $\triangle ABC$ is an isosceles right angled triangle, and (S2): the area of $\triangle ABC$ is $\dfrac{9\sqrt{2}}{2}$

Let P be the foot of the perpendicular from the point Q(10, -3, -1) on the line x-3 y-2 z+1 7 = -1 = -2 . Then the area of the right angled triangle P QR, where R is the point (3, -2, 1), is

The lines $\dfrac{x-2}{2}=\dfrac{y}{-2}=\dfrac{z-7}{16}$ and $\dfrac{x+3}{4}=\dfrac{y+2}{3}=\dfrac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z-1}{1}$ is $l$, then $14l^2$ is equal to

If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line $x - 3y + 2z - 1 = 0$, $4x - y + z = 0$ is $Ax + By + Cz = 1$, then $140(C - B + A)$ is equal to _____.

Let a straight line L pass through the point P (2, -1, 3) and be perpendicular to the lines y+1 x-1 2 = 1 = z-3 -2 and x-3 y-2 z+2 1 = 3 = 4 . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is :

Let the co-ordinates of one vertex of $\triangle ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. For $\alpha \in \mathbb{Z}$, if the area of $\triangle ABC$ is 21 sq. units and the line segment BC has length $2\sqrt{21}$ units, then $\alpha^2$ is equal to _____.

Let L : y-1 y z+4 1 x-1 3 = -1 = z+1 0 and L : 2 x-2 2 = 0 = \alpha ,\alpha \in 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 , then the value of 26\alpha( PB) is _________ 2 2

Image of $P(1,2,3)$ in plane $2x-y+z=9$ is $Q$. $R=(6,10,7)$. Then square of area of $\triangle PQR$ is _______

Ex. 61 (A): If the line \(\frac{x-1}{1} = \frac{y+1}{-2} = \frac{z+1}{\lambda}\) lies in the plane \(3x - 2y + 5z = 0\), then \(\lambda\) is equal to

$P$ = intersection of $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ with $x+y+z=2$. Distance from $P$ to $3x-4y+12z=32$ is $q$. Then $q$ and $2q$ are roots of

Let $L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $L_2:\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

The equation of a plane is \(x + 2y + 2z + 7 = 0\) and the equation of a sphere is \(x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0\). The radius of the circle of intersection of the plane and sphere is:

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

If an angle between the line, \(\dfrac{x+1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{-2}\) and the plane, \(x - 2y - kz = 3\) is \(\cos^{-1}(2\sqrt{2}/3)\), then a value of \(k\) is ______ (up to three decimal places).

Let a unit vector $\overrightarrow{OP}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the co-ordinate axes OX, OY, OZ respectively, where $\beta \in \left(0, \frac{\pi}{2}\right)$. $\overrightarrow{OP}$ is perpendicular to the plane through points $(1,2,3)$, $(2,3,4)$ and $(1,5,7)$. Then which one of the following is true?

Find the length of the projection of vector $\vec{AB}$ onto the normal vector to a plane, where $A = (1, 2, -1)$, $B = (3, 5, 5)$, and the normal vector is $3\vec{i} - 4\vec{j} + 12\vec{k}$.

Line $L$ through $(0,1,2)$ intersects $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and is parallel to $2x+y-3z=4$. Distance of $P(1,-9,2)$ from $L$ is

Image of $P(1,2,6)$ in plane through $A(1,2,0)$, $B(1,4,1)$, $C(0,5,1)$ is $Q(\alpha,\beta,\gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to

Let the plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$. If the intercept of P on the y-axis is 1, then the distance between P and L is:

Let the plane P pass through the intersection of the planes $2x + 3y - z = 2$ and $x + 2y + 3z = 6$, and be perpendicular to the plane $2x + y - z + 1 = 0$. If $d$ is the distance of P from the point $(-7, 1, 1)$, then $d^2$ is equal to:

Let a line parallel to \(z\)-axis passing through a point \(P(3, 4, a)\) intersects the plane \(x - 2y + 2z = a^2 + 4a + 1\) at \(Q\) where \(a \in R\). If least area of \(\triangle OPQ\) is equal to \(\left(\dfrac{p}{q}\right)\) where \(p\) and \(q\) are co-prime numbers, then find the value of \((p + q)\).

Plane containing $x+2y+3z-4=0=2x+y-z+5$ and $\perp$ to $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})$ is $ax+by+cz=4$. Then $a-b+c$ is equal to

Let $PQR$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the midpoint of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the lines $\dfrac{x-2}{0}=\dfrac{y}{2}=\dfrac{z+3}{-1}$ and $\dfrac{x-1}{1}=\dfrac{y+3}{-3}=\dfrac{z+1}{1}$ is

$L$ through origin $\perp L_1$ and $L_2$. $P$ = intersection of $L$ and $L_1$. $Q(\alpha,\beta,\gamma)$ = foot of $\perp$ from $P$ on $L_2$. Then $9(\alpha+\beta+\gamma)$ is equal to ________.

The distance, of the point $(7,-2,11)$ from the line $\dfrac{x-6}{1}=\dfrac{y-4}{0}=\dfrac{z-8}{-3}$ along the line $\dfrac{x-5}{2}=\dfrac{y-1}{-3}=\dfrac{z-5}{6}$, is:

Plane through $(0,-1,2)$ and $(-1,2,1)$, parallel to line through $(5,1,-7)$ and $(1,-1,-1)$. Which point lies on it?

A line with direction ratios $2,1,2$ meets the lines $x=y+2=z$ and $x+2=2y=2z$ respectively at the points $P$ and $Q$. If the length of the perpendicular from the point $(1,2,12)$ to the line $PQ$ is $l$, then $l^2$ is

The shortest distance between the lines $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}$ and $\frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$ is equal to _______.

Plane $P$ contains line of intersection of $\vec{r}\cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r}\cdot(2\hat{i}+3\hat{j}+4\hat{k})=-5$. $P$ passes through $(0,2,-2)$. Square of distance from $(12,12,18)$ to $P$ is

A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60°$, to the y-axis at $45°$ and to the z-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$, is normal to $\vec{v}$, then

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

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{-2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{3}=\dfrac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ is:

Normal vector to the plane $P = 0$ is $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 2 \\ 3 & 6 & -2 \end{vmatrix} = \vec{i}(-10) - \vec{j}(-10) + \vec{k}(15) = -10\vec{i} + 10\vec{j} + 15\vec{k}$. Equation of the plane through $(1, 1, 1)$ is $-2(x - 1) + 2(y - 1) + 3(z - 1) = 0$. So, the distance from the point $(1, 2, 3)$ is $\left|\frac{-1 + 3 - 3}{\sqrt{17}}\right| = \frac{1}{\sqrt{17}}$ units.

Let L : y-2 y-4 and L : be two lines. Then which of the following points lies on x-1 z-3 x-2 z-5 1 = = 2 = = 2 3 4 3 4 5 the line of the shortest distance between L and L ? 1 2

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x + 2y + z = 0$ and $3y - z = 3$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to

The point of intersection C of the plane $8x + y + 2z = 0$ and the line joining the points $A(-3, -6, 1)$ and $B(2, 4, -3)$ divides the line segment AB internally in the ratio $k:1$. If $a, b, c$ ($|a|, |b|, |c|$ are coprime) are the direction ratios of the perpendicular from the point C on the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$, then $|a+b+c|$ is equal to _____.

$P_1:\ 3x-y-7z=11$; $P_2$ through $(2,-1,0),(2,0,-1),(5,1,1)$. Foot of perpendicular from $(7,4,-1)$ on $P_1\cap P_2$ is $(\alpha,\beta,\gamma)$. Then $\alpha+\beta+\gamma$ is equal to

Plane $4x-y+z=10$ rotated by $\frac{\pi}{2}$ about its intersection with $x+y-z=4$. $\alpha$ = distance of $(2,3,-4)$ from new plane. $35\alpha$ is equal to

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

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point P. Let Q be the foot of perpendicular from the point $R(1, -1, -3)$ on the line L. If $\alpha$ is the area of triangle PQR, then $\alpha^2$ is equal to _____.

Let the equation of the plane passing through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ and parallel to the line $x + y + 2z - 7 = 0 = 2x + 3y + z - 2$ be $ax + by + cz = 65$. Then the distance of the point $(a, b, c)$ from the plane $2x + 2y - z + 16 = 0$ is ______.

Plane through intersection of $x+2y+az=2$ and $x-y+z=3$ is $5x-11y+bz=6a-1$. For $c\in\mathbb{Z}$, distance from $(a,-c,c)$ is $\frac{2}{\sqrt{a}}$. Then $a+b$ is equal to

Find the shortest distance between two skew lines given position vectors and magnitude conditions for a tetrahedron $ABCD$.