3D Geometry Questions - JEE

In a three dimensional coordinate system P, Q and R are images of a point A(a, b, c) in the XY, YZ and the ZX planes respectively. If G is the centroid of triangle PQR, then area of triangle AOG is (O is the origin)

If the plane $2ax - 3ay + 4az + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres $x^2 + y^2 + z^2 + 6x - 8y - 2z = 13$ and $x^2 + y^2 + z^2 - 10x + 4y - 2z = 8$, then a equals

The line of intersection of the planes, $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$, is

Lines are $\dfrac{x}{0} = \dfrac{y}{0} = \dfrac{z}{1} = \lambda$ (z axis) and $x + y + 2z - 3 = 0,\ 2x + 3y + 4z - 4 = 0$. The shortest distance (S.D.) between the two lines is given by $\text{S.D.} = \dfrac{|(\vec{c}-\vec{a})\cdot(\vec{b}\times\vec{d})|}{|\vec{b}\times\vec{d}|}$. Find the shortest distance.

A variable plane at a distance of 1 unit from the origin cuts the coordinate axes at A, B and C. If the centroid D(x, y, z) of triangle ABC satisfies the relation $\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} = k$, then the value of k is __________.

The sum of the intercepts on the coordinate axes of the plane passing through the point (–2, –2, 2) and containing the line joining the points (1, –1, 2) and (1, 1, 1), is

The shortest distance between the lines $\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}$ and $\dfrac{x+2}{-1} = \dfrac{y-4}{8} = \dfrac{z-5}{4}$ lies in the interval

The coordinates of a point on the plane 2x + y - 5z = 0, which is $$2\sqrt{11}$$ units away from the line of intersection of 2x + y - 5z = 0 and 4x - 3y + 7z = 0 are:

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:

The equation of the plane containing the line $2x - 5y + z = 3$; $x + y + 4z = 5$, and parallel to the plane, $x + 3y + 6z = 1$, is

A line $AB$ in three-dimensional space makes angles $45°$ and $120°$ with the positive $x$-axis and the positive $y$-axis, respectively. If $AB$ makes an acute angle $\theta$ with the positive $z$-axis, then $\theta$ equals

If the image of the point $P(1, -2, 3)$ in the plane, $2x + 3y - 4z + 22 = 0$ measured parallel to the line, $\dfrac{x}{1} = \dfrac{y}{4} = \dfrac{z}{5}$ is $Q$, then $PQ$ is equal to

The point of intersection of the plane $\vec{r} \cdot (3\hat{i} - 5\hat{j} + 2\hat{k}) = 6$ with the straight line passing through the origin and perpendicular to the plane $2x - y - z = 4$, is $(x_0, y_0, z_0)$. The value of $(2x_0 - 3y_0 + z_0)$, is:

Let P(x, y, z) be any point on the locus, then the distances from the six faces are |x + 1|, |x − 1|, |y + 1|, |y − 1|, |z + 1| and |z − 1|. According to the given condition, find the locus of P.$|x+1|^2 + |x-1|^2 + |y+1|^2 + |y-1|^2 + |z+1|^2 + |z-1|^2 = 10$

83. We have three planes$P_1: 2x - y + 2z + 3 = 0$$P_2: 2x - y + 2z + \lambda/2 = 0$$P_3: 2x - y + 2z + \mu = 0$Given that the distance between $P_1$ and $P_2$ is $\frac{1}{3}$ and the distance between $P_1$ and $P_3$ is $\frac{2}{3}$. Find $\lambda_{\max} + \mu_{\max}$.

Given lines are: $\vec{r} = (\hat{i} + (-3\hat{j}) + \hat{k}) + s(\hat{i} - \lambda\hat{j} + \lambda\hat{k})$ and $\vec{r} = (0\hat{i} + \hat{j} + 2\hat{k}) + t\left(\dfrac{1}{2}\hat{i} + \hat{j} - \hat{k}\right)$. If $\vec{a} = (1-0)\hat{i} + (-3-1)\hat{j} + (1-2)\hat{k}$, $\vec{b} = \hat{i} - \lambda\hat{j} + \lambda\hat{k}$, and $\vec{c} = \dfrac{1}{2}\hat{i} + \hat{j} - \hat{k}$ are coplanar, then $\lambda$ equals:

Given planes $3x + 4y + z = 1$ and $5x + 8y + 2z + 14 = 0$. The sine of the angle between the plane $x + y + z = 5$ and the line of intersection of the two given planes is:

The lines $\dfrac{x-2}{1} = \dfrac{y-3}{1} = \dfrac{z-4}{-k}$ and $\dfrac{x-1}{k} = \dfrac{y-4}{2} = \dfrac{z-5}{1}$ are coplanar is

Equation of the line through the point (1, 1, 1) and intersecting the lines $2x - y - z - 2 = 0 = x + y + z - 1$ and $x - y - z - 3 = 0 = 2x + 4y - z - 4$

The point of intersection of the plane $\vec{r} \cdot (3\hat{i} - 5\hat{j} + 2\hat{k}) = 6$ with the straight line passing through the origin and perpendicular to the plane $2x - y - z = 4$, is $(x_0, y_0, z_0)$. The value of $(2x_0 - 3y_0 + z_0)$ is:

A plane containing the point $(3, 2, 0)$ and the line $\dfrac{x-1}{1} = \dfrac{y-2}{5} = \dfrac{z-3}{4}$ also contains the point

A plane passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane where $a, b, c$ (\(a

If the straight lines $x = 1 + s,\ y = -3 - \lambda s,\ z = 1 + \lambda s$ and $x = \dfrac{t}{2},\ y = 1 + t,\ z = 2 - t$ with parameters $s$ and $t$, respectively, are co-planar then $\lambda$ equals

A variable plane passes through a fixed point $(3, 2, 1)$ and meets $x$, $y$ and $z$ axes at $A$, $B$ and $C$, respectively. A plane is drawn parallel to $yz$-plane through $A$, a second plane is drawn parallel to $zx$-plane through $B$ and a third plane is drawn parallel to $xy$-plane through $C$. Then the locus of the point of intersection of these three planes, is

If direction cosines of a line are $\langle l, m, n \rangle$ such that $l^2 + m^2 + n^2 = 1$ and $\alpha = \theta$, $\beta = \beta$, $\gamma = \theta$, then $\cos^2\theta$ equals:

The equation of the plane that passes through the points (1, 1, 0), (1, 2, 1), and (−2, 2, −1) is

Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle α with the positive X-axis, then cos α is equal to

The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is

The two lines x = ay + b, z = cy + d and x = a'y + b', z = c'y + d' are perpendicular to each other if

Image of point Q in a plane is found using: $\dfrac{x-0}{3} = \dfrac{y+1}{-1} = \dfrac{z+3}{4} = \dfrac{-2(1-12-2)}{9+1+16} = 1$. Given points P(3, -2, 1), Q(0, -1, -3) and R(3, -1, -2), find the area of triangle PQR (in square units, rounded to 3 decimal places).

The angle between a line with direction ratios proportional to $2, 2, 1$ and a line joining $(3, 1, 4)$ to $(7, 2, 12)$ is:

We have the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$.The shortest distance between these two lines lies in the interval:

The projection of a line segment on the coordinate axes are 2, 3, 6. Then, the length of the line segment is

A plane 2x + 3y + 5z = 1 has a point P which is at minimum distance from line joining A(1, 0, -3), B(1, -5, 7), then distance AP is equal to

Equation of plane containing both lines is:

The length of the perpendicular from vertex D on the opposite face is

The intersection of the spheres $x^2 + y^2 + z^2 + 7x - 2y - z = 13$ and $x^2 + y^2 + z^2 - 3x + 3y + 4z = 8$ is the same as the intersection of one of the sphere and the plane

Two parallel planes are given by $x + y + z = 1$ and $x + y + z = \dfrac{9}{2}$. A third plane that intersects them is given by $2x - 5y + z = -5$, resulting in two parallel lines of intersection. If the distance $d$ between these two parallel lines can be expressed as $d = \sqrt{\dfrac{a}{b}}$, where $a$ and $b$ are co-prime positive integers, then find the value of $[d]$.[Note: Where $[k]$ denotes greatest integer function less than or equal to $k$.]

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\dfrac{x}{1}=\dfrac{1-y}{2}=\dfrac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1,6,3)$ and $(4,9,\alpha)$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $\triangle ABC$ is:

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

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

The angle between the line $\frac{x + 1}{2} = \frac{y}{3} = \frac{z - 3}{6}$ and the plane $10x + 2y - 11z = 3$ is

If lines $x = y = z$, $y = \frac{z}{2} = \frac{3}{3}$ and the third line passing through $(1, 1, 1)$ form a triangle of area $\sqrt{6}$ units, then point of intersection of third line with second line will lie on:

The direction cosines of the shortest distance lie between the planes $y + z = 0$ and $z + x = 0$ is:

The plane which bisects the line segment joining the points (−3, −3, 4) and (3, 7, 6) at right angles passes through which one of the following points?

Let $L_1:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda(\hat{i}-\hat{j}+2\hat{k})$, $L_2:\vec{r}=(\hat{j}-\hat{k})+\mu(3\hat{i}+\hat{j}+p\hat{k})$ and $L_3:\vec{r}=\delta(l\hat{i}+m\hat{j}+n\hat{k})$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

Consider the plane through $(2, 3, -1)$ and at right angles to the vector $3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}$ from the origin is

Radius of the sphere, with (2, -3, 4) and (-5, 6, -7) as extremities of a diameter, is

The coordinates of the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the join of the points $B(4, 7, 1)$ and $C(3, 5, 3)$ are

If the lines $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$ and $\dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1}$ intersect, then $k$ is equal to

3D Geometry Questions (578)