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

A unit vector parallel to the straight line $\vec{r} = -(5 + 4s)\hat{i} + (7 - 2s)\hat{j} + (3 + 4s)\hat{k}$, where $s$ is the parameter of the line, is

If the equation of the straight line passing through the point \((a,1,3)\) and parallel to the vector \(\frac{2}{3}\hat{i} + \frac{3}{2}\hat{j} + \hat{k}\) is \(\frac{3x+6}{b} = \frac{2y-2}{3} = \frac{z-3}{1}\), then the value of \(a+b\) is equal to

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:

Which one of the following is a vector parallel to the straight line $\vec{r}=(\hat{i}-11\hat{j}+101\hat{k})+\lambda(3\hat{i}-5\hat{j}+2\hat{k}),\lambda\in\mathbb{R}$? ________.

The equation of the line passing through (0, 0, 1) and (1, 1, 0) is ________.

The point of intersection of the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-11}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ is ________.

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 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

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

The point of intersection of the straight lines $\vec{r} = (3\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(-\hat{i} - 2\hat{j} + 2\hat{k})$ and $\frac{3-x}{-1} = \frac{y+4}{2} = \frac{z-5}{7}$ is:

The vector equation of the straight line $\frac{x-2}{1} = \frac{y}{-3} = \frac{1-z}{2}$ is:

The straight line $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + \alpha(2\hat{i} - \hat{j} + 4\hat{k})$ meets the $xy$-plane at the point:

The direction cosines of the straight line given by the planes $x=0$ and $z=0$ are: