3D Geometry Questions (578)

Find the image of point A(2, 1, 6) about the mirror plane x + y - 2z = 3.

Ex. 45 Statement I: Line \(\frac{x-1}{3} = \frac{y-2}{11} = \frac{z+1}{11}\) lies in the plane \(11x - 3z - 14 = 0\).Statement II: A straight line lies in a plane, if the line is parallel to the plane and a point of the line is in the plane.

Given the planes x + 2y − 3z + 2 = 0 and x − 2y + 3z + 7 = 0, if the point P is (1, 2, 2), then

For positive l, m and n, if the planes x = ny + mz, y = lz + nx, z = mz + ly intersect in a straight line, then l, m and n satisfy the equation

Through a point \(P(h, k, l)\) a plane is drawn at right angles to OP to meet the coordinate axes in A, B and C. If \(OP = p\), \(A_{xy}\) is area of projection of \(\triangle ABC\) on xy-plane, \(A_{yz}\) is area of projection of \(\triangle ABC\) on yz-plane, then \(\frac{A_{xy}}{A_{yz}}\)

Let $d$ be the distance of the point of intersection of the lines $\dfrac{x+6}{3}=\dfrac{y}{2}=\dfrac{z+1}{1}$ and $\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}$ from the point $(7,8,9)$. Then $d^2+6$ is equal to:

$P$ is a point on the plane $lx + my + nz = p$. A point $Q$ is taken on the line $OP$ such that $OP.OQ = p^2$. Then the locus of $Q$ is:

The orthogonal projection of the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-1}{4}$ on the plane $3x + 4y + 5z = 0$ is:

If $abc \neq 0$ and let $(p_1, q_1, r_1)$ be the image of $(p, q, r)$ in the plane $ax + by + cz + d = 0$, then:

The foot of the perpendicular from the point $O(0, 0, 0)$ to the line of intersection of the planes $x + y + z = 4$ and $2x + y + 3z = 1$ is point $A$. Then the equation of line $OA$ is:

A variable line passing through the point $P(0, 0, 2)$ always makes angle $60°$ with $z$-axis, intersects the plane $x + y + z = 1$. Then the locus of point of intersection of the line and the plane is:

The locus of intersection of locus of $P$ with the plane $x + y + z = 1$ is:

A mirror and a source of light are situated at the origin $O$ and at a point on the line $OX$ respectively. A ray of light from the source strikes the mirror at $O$ and is reflected. If the direction ratios of the normal to the plane of the mirror are $(1, -1, 1)$; then the direction cosines of the reflected ray are :

Given \(\overrightarrow{OQ} = (1-3\mu)\hat{i} + (\mu-1)\hat{j} + (5\mu+2)\hat{k}\) and \(\overrightarrow{OP} = 3\hat{i} + 2\hat{j} + 6\hat{k}\) (where O is the origin). If \(\overrightarrow{PQ}\) is parallel to the plane \(x - 4y + 3z = 1\), find the value of \(\mu\).

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to:

If the line $x = y = z$ intersect the line in $A x + \sin B y + \sin C z = 2d^2$, sin $2A x + \sin 2B y + \sin 2C z = d^2$ then $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ is equal to : (where $A + B + C = \pi$)

The equation of a line in $xz$ plane equally inclined with $x$ and $z$ axes which is at a unit distance from the line $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z}{1}$ is:

The coordinate of the points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ which are at a distance $3\sqrt{2}$ from the point $(1, 2, 3)$

The equation of the plane parallel to plane $x + y + 2z = 5$ at a distance $\sqrt{6}$ units from the plane is/are:

The shortest distance between the lines $2x + y + z = 1, 3x + y + 2z = 2$ and $x + y = z$ is $d$ then $\frac{1}{d^2} = $ ______.

Three lines $y - z - 1 = 0, x = 0; x + z - 1 = 0, y = 0; x - z - 1 = 0, y = 0$ intersect the $xy$ plane at $A, B$ and $C$. If the orthocentre of $\triangle ABC$ is $(p, q, r)$ then $3p + q + r = $ __________.

The minimum distance of the point $(1, 1, 1)$ from the plane $x + y + z = 1$ measured perpendicular to the line $\frac{x-x_1}{l} = \frac{y-y_1}{2} = \frac{z-z_1}{3}$ is then $\frac{3\ell^2}{\gamma} = $ __________.

The maximum distance between the point $P(0, 0, 3)$ and the circle $x^2 + y^2 - 2\sqrt{5}x - 4y + 8 = 0; z = 0$ is __________.

Plane $2x + 3y + 4z = 5$ is rotated about the line where it cuts the $xy$-plane by an angle $\alpha$. In the new position the plane contains the point $(1, 1, 1)$. If the angle $\alpha = \cos^{-1}\sqrt{\frac{p}{q}}, (p$ is rational number in its simplest form$)$ then $q - 2p = $ __________.

If $a, b, c$ be the lengths of the intercepts of the plane passing through the intersection of the planes $2x + y + 2z = 9, 4x - 5y - 4z = 1$ and the point $(3, 2, 1)$ on the coordinate axes, then $(5a + b + c)/2 = $ __________.

If $Q$ is the foot of perpendicular from the point $P(4, -5, 3)$ on the line $\frac{x-5}{3} = \frac{y+2}{-4} = \frac{z-6}{5}$, then $[PQ] = $ __________. (where $[.]$ denote greatest integer function)

The projection of the line $\frac{x}{5} = \frac{y-1}{2} = \frac{z-1}{1}$ on a plane $P$ is $\frac{x}{1} = \frac{y-1}{1} = \frac{z-1}{-1}$. The plane $P$ passes through $(k, -2, 0)$ then $k = $ __________.

Let for $\lambda \in [0, \infty)$ such that $(x, y, z) \neq (0, 0, 0)$ and $(\vec{i} + \vec{j} + 3\vec{k})x + (3\vec{i} - \vec{j} + \vec{k})y + (4\vec{i} + 5\vec{j})z = \lambda(x\vec{i} + y\vec{j} + 3\vec{k})$, then the value of $\frac{x - y - z}{x}$ is equal to __________.

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 ______.