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

The point at which the line $\frac{x-2}{1}=\frac{y-4}{-5}=\frac{z+3}{4}$ intersects the xy-plane is

The Cartesian equation of the line passing through the points $(1,-1,2)$ and $(7, 0, 5)$ is

Which one of the following points lies on the straight line $\frac{x-1}{2}=\frac{y+1}{4}=\frac{z-2}{-2}$?

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:

The foot of the perpendicular from the point \( (1,6,3) \) to the line \( \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \) is

The projection of the line segment joining \( (2,0,-3) \) and \( (5,-1,2) \) on a straight line whose direction ratios are \( 2,4,4 \) is equal to

A unit vector parallel to the straight line \( \frac{x - 2}{3} = \frac{3 + y}{-1} = \frac{z - 2}{-4} \) is:

A unit vector parallel to the straight line \( \frac{x - 2}{3} = \frac{3 + y}{-1} = \frac{z - 2}{-4} \) is:

A unit vector parallel to the straight line \( \frac{x - 2}{3} = \frac{3 + y}{-1} = \frac{z - 2}{-4} \) is:

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