List of top Mathematics Questions on Equation of a Line in Space

The lines \( \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \) and \( \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \) intersect at a point, where \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = \hat{i} - \hat{k} \). Find the point of intersection.

The vector form of the straight line $\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z-1}{-2}$ is

The equation of line which is parallel to $\frac{2-x}{-3}=\frac{y-2}{2}=\frac{z-4}{1}$ and passing through the point $(1,1,1)$, is

Consider the straight line $\vec{r} = (5\hat{i} + 2\hat{j} - 3\hat{k}) + t(4\hat{i} + 6\hat{j} - 7\hat{k}), \; t \in \mathbb{R}$. Which one of the following points is a point on the straight line?

A straight line passes through the point whose position vector is $\hat{k}$. The straight line also passes through the point of intersection of the lines $\vec{r} = \hat{j} + \lambda \hat{i}, \lambda \in \mathbb{R}$ and $\vec{r} = \hat{i} + s\hat{j}, s \in \mathbb{R}$. Then the equation of the straight line is:

The equation of a line passing through $(-1,2,-4)$ and parallel to the straight line $\dfrac{-x-1}{4} = \dfrac{2y+1}{-1} = \dfrac{-z+4}{3}$, is:

Identify the co-ordinates of the point where the line joining \( (1,1,1) \) and \( (2,2,2) \) intersects the plane \( x+y+z=9 \).

The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is

The number of solutions of \(16^{\sin^2 x} + 16^{\cos^2 x} = 10\) in \(0 \le x \le 2\pi\) are

If the point \( (3,6,k) \) lies on the line \( \dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3} \), then the value of \( k \) is

If a point \( P \) with \( x \)-coordinate \( 7 \) lies on the line joining the points \( A(1,2,3) \) and \( B(4,6,8) \), then the coordinates of the point \( P \) are

The equation of the straight line joining the points \((1,2,3)\) and \((3,4,k)\) is \(\frac{x-3}{1}=\frac{y-4}{1}=\frac{z-k}{5}\). Then the value of \(k\) is

The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is

The point at which the line $\frac{x+3}{11}=\frac{y-2}{-1}=\frac{z+1}{3}$ meets the $zx$-plane is:

Which one of the following is a point on the straight line $\vec{r}=(13\hat{i}-14\hat{j}+23\hat{k})+\lambda(5\hat{i}-7\hat{j}-9\hat{k})$?

The Cartesian equation of the line $\vec{r}=(2\hat{i}-7\hat{j}+11\hat{k})+\lambda(3\hat{i}+7\hat{j}-13\hat{k})$ is:

Let $\vec{OP}=2\hat{j}$ be the position vector of a point $P$. Let $\vec{r}=\hat{j}+\lambda(\hat{i}+\hat{j})$ be a straight line. The distance of the point $P$ from the line is:

A straight line passes through the points \( (10,8,6) \) and \( (13,9,4) \). A unit vector parallel to this line is

A straight line passing through \( (6,1,3) \) meets the line \( \frac{x-1}{2} = \frac{y}{1} = \frac{z-2}{3} \) at \( Q \). If the lines are perpendicular to each other, then the coordinates of \( Q \) are

Two lines \( \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} \) and \( \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} \) intersect at a point. Then the value of \( k \) is

The equation of a line passing through origin with direction angles \( \frac{2\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3} \) is

The image of a point \( P(3,5,3) \) in the line \( \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \) is \( P'(a,b,c) \). Then \( a+b+c = \)

The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is

If the line $\dfrac{x+1}{4} = \dfrac{y+2}{-3} = \dfrac{z-\alpha}{-2}$ passes through the point $(-1,\,-2,\,-3)$, then the value of $\alpha$ is

The line $\dfrac{x+1}{2} = \dfrac{y-4}{4} = \dfrac{z-2}{5}$ passes through the point