3D Geometry Questions - JEE

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

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

The equation of a plane is $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$. The plane passes through the point $(3, 2, 1)$ and meets the axes at $A(a, 0, 0)$, $B(0, b, 0)$ and $C(0, 0, c)$. The locus of the point of intersection of planes through $A$, $B$ and $C$ parallel to the $yz$-, $zx$- and $xy$-planes respectively is:

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$, $(1, -\lambda^2, 1)$ and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to:

A line segment has direction cosines $l, m, n$. If the line makes an angle of $45°$ with the $x$-axis and $120°$ with the $y$-axis, then the angle made by the line with the positive $z$-axis is:

Let the lines $L_1:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ and $L_2:\vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k})$ intersect at the point $R$. Let $P$ and $Q$ be points on $L_1$ and $L_2$ respectively such that $|\overrightarrow{PR}|=\sqrt{29}$ and $|\overrightarrow{PQ}|=\sqrt{47/3}$. If $P$ lies in the first octant, then $27(QR)^2$ is equal to

The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$ is

If $10y - 8x - (x^2 + y^2 + z^2) = 40, P_1 = \max\left\{\sqrt{(x+2)^2 + (y-3)^2 + z^2}\right\}, P_2 = \min\left\{\sqrt{(x+2)^2 + (y-3)^2 + z^2}\right\}$, then $P_1 - P_2$ is __________.

The perpendicular distance from the origin to the plane containing the two lines, $\dfrac{x+2}{3} = \dfrac{y-2}{5} = \dfrac{z+5}{7}$ and $\dfrac{x-1}{1} = \dfrac{y-4}{4} = \dfrac{z+4}{7}$, is ______ (up to three decimal places).

If the direction cosines of two lines are such that l + m + n = 0 and l^2 + m^2 - n^2 = 0, then the angle between them is

If the length of the perpendicular from the point $(\beta, 0, \beta)$ ($\beta \neq 0$) to the line, $\dfrac{x}{1} = \dfrac{y-1}{0} = \dfrac{z+1}{-1}$ is $\sqrt{\dfrac{3}{2}}$, then $|\beta|$ is equal to ______.

The vector equation of plane which is at a distance of 8 units from the origin and which is normal to the vector $2\vec{i} + \vec{j} + 2\vec{k}$ is $\vec{r} \times (2\vec{i} + \vec{j} + 2\vec{k}) = l$, where $l$ is equal to

The equation of a line of greatest slope on an inclined plane can be represented as:

If $\cos^2\theta + \cos^2\theta + \cos^2\alpha = 1$ (sum of squares of direction cosines = 1), find the range of $\theta$ given that $0 \leq 2\cos^2\theta \leq \frac{1}{2}$.

The shortest distance between the lines x + 1, y + 1, z + 1 is

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y = 15$, then $2\alpha - 3\beta$ is equal to ______.

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

Equation of plane passing through the points C_1(2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z - 1 = 0, is

If direction ratios of a line are $l, m, n$ and the direction ratios of normal to a plane are $a, b, c$, then the condition for the line to be parallel to the plane is:

The distance of the point $Q(0,2,-2)$ from the line passing through the point $P(5,-4,3)$ and perpendicular to the lines $\vec{r}=(-3\hat{i}+2\hat{k})+\lambda(2\hat{i}+3\hat{j}+5\hat{k})$ and $\vec{r}=(\hat{i}-2\hat{j}+\hat{k})+\mu(-\hat{i}+3\hat{j}+2\hat{k})$ is

Equation of the line of the shortest distance between the lines $\dfrac{x}{1} = \dfrac{y}{-1} = \dfrac{z}{1}$ and $\dfrac{x-1}{0} = \dfrac{y+1}{-2} = \dfrac{z}{1}$ is

The plane containing the line $\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z-3}{3}$ and parallel to the line $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{1}$ passes through the point

The shortest distance between lines $L_1$ and $L_2$, where $L_1:\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4,4,3)$, $B(-1,6,3)$ and perpendicular to the line $\dfrac{x-3}{-2}=\dfrac{y}{3}=\dfrac{z-1}{1}$, is

3D Geometry Questions (578)