Q.Find the Cartesian equation of the line passing through the point $(1,2,-4)$ and perpendicular to the two lines $\dfrac{x-8}{3} = \dfrac{y+19}{-16} = \dfrac{z-10}{7}$ and $\dfrac{x-15}{3} = \dfrac{y-29}{8} = \dfrac{z-5}{-5}$. Also write the vector form of the line so obtained. (5+1) OR Find the shortest distance between the lines whose vector equations are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$

Line Perpendicular to Two Lines

In 3D geometry a very common task is this: two lines are given, and you must find the direction of a third line that is perpendicular to both of them. This shows up when finding a common perpendicular, the shortest distance between two lines, or the normal to a plane containing two directions.

The Core Idea

A line in space is fixed by two things: a point it passes through, and its direction vector. So "find a line perpendicular to two lines" really means "find a vector perpendicular to both of the two given direction vectors."

If the two given lines have direction vectors

then a vector perpendicular to both is their cross product:

Why the Cross Product?

The defining property of $\vec{b_1}\times\vec{b_2}$ is that it is perpendicular to each factor:

So it points in exactly the direction we need — along both perpendicularity conditions at once. This is why one cross product replaces solving a pair of dot-product equations by hand.

Using It

Suppose a line must be perpendicular to $\vec{b_1} = \hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b_2}=\hat{i}-\hat{j}+\hat{k}$.

So the required line has direction ratios $\langle 5, 2, -3\rangle$. Through a point $A(x_0,y_0,z_0)$ its equation is …