List of top Mathematics Questions on 3D Geometry

If the line \(\frac{2 - x}{3} = \frac{3y - 2}{4\lambda + 1} = 4 - z\) makes a right angle with the line  \(\frac{x + 3}{3\mu} = \frac{1 - 2y}{6} = \frac{5 - z}{7},\) then \( 4\lambda + 9\mu \) is equal to:

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

Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If ∠QPR is a right angle, then 12a² is equal to _____

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 \[ \vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 5\hat{k}), \quad \lambda \in \mathbb{R} \] and \[ \vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu (-\hat{i} + 3\hat{j} + 2\hat{k}), \quad \mu \in \mathbb{R} \] is

A line passes through \( A(4, -6, -2) \) and \( B(16, -2, 4) \). The point \( P(a, b, c) \) where \( a \), \( b \), \( c \) are non-negative integers, on the line \( AB \) lies at a distance of 21 units from the point \( A \). The distance between the points \( P(a, b, c) \) and \( Q(4, -12, 3) \) is equal to \(\_\_\_\_\_\).

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

The square of the distance of the image of the point \( (6, 1, 5) \) in the line \[ \frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}, \] from the origin is _____ .

Let \( (\alpha, \beta, \gamma) \) be the mirror image of the point \( (2, 3, 5) \) in the line \(\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}\). Then \( 2\alpha + 3\beta + 4\gamma \) is equal to

Consider the line \( L \) passing through the points \( (1, 2, 3) \) and \( (2, 3, 5) \). The distance of the point \[ \left( \frac{11}{3}, \frac{11}{3}, \frac{19}{3} \right) \] from the line \( L \) along the line \[ \frac{3x - 11}{2} = \frac{3y - 11}{1} = \frac{3z - 19}{2} \] is equal to:

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

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 a distance of 9 units from the point \( P \), be \( (\alpha, \beta, \gamma) \). Then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

Let a line passing through the point \((-1, 2, 3)\) intersect the lines\( L_1 : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{-2} \) at \( M(\alpha, \beta, \gamma) \) and\( L_2 : \frac{x + 2}{-3} = \frac{y - 2}{-2} = \frac{z - 1}{4} \) at \( N(a, b, c) \).Then the value of \( \frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} \) equals ____.

The angle between two lines whose direction ratios are proportional to \(1, 1, -2\) and \((\sqrt{3} - 1), (-\sqrt{3} - 1), -4\) is:

The direction cosines of the line which is perpendicular to the lines with direction ratios 1,-2,-2 and 0, 2, 1 are:

Let \( P(3, 2, 3) \), \( Q(4, 6, 2) \), and \( R(7, 3, 2) \) be the vertices of \( \triangle PQR \). Then, the angle \( \angle QPR \) is

Let \( L_1 : \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda (\hat{i} - \hat{j} + 2\hat{k}) \), \( \lambda \in \mathbb{R} \)
\( L_2 : \vec{r} = (\hat{j} - \hat{k}) + \mu (3\hat{i} + \hat{j} + p\hat{k}) \), \( \mu \in \mathbb{R} \) and
\( L_3 : \vec{r} = \delta (\ell \hat{i} + m \hat{j} + n \hat{k}) \), \( \delta \in \mathbb{R} \)
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

If \( d_1 \) is the shortest distance between the lines  \(\frac{x+1}{2} = \frac{y-1}{-12} = \frac{z-1}{-7} \quad \text{and} \quad \frac{x-1}{7} = \frac{y+8}{2} = \frac{z-4}{5},\) and \( d_2 \) is the shortest distance between the lines \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3} \quad \text{and} \quad \frac{x+2}{1} = \frac{y+2}{2} = \frac{z-1}{6},\) then the value of \( \frac{32\sqrt{3}d_1}{d_2} \) is:

If the shortest distance between the lines \(\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{-3} and \frac{x - \lambda}{2} = \frac{y + 1}{4} = \frac{z - 2}{-5}\) is \(\frac{6}{\sqrt{5}}\), then the sum of all possible values of \(\lambda\) is:

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

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

Let \( O \) be the origin, and \( M \) and \( N \) be the points on the lines \[\frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3} \]and\[\frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9} \]respectively, such that \( MN \) is the shortest distance between the given lines. Then \( OM \cdot ON \) is equal to ______.