Q.Find the equation of the line which bisects the line segment joining points $A(2, 3, 4)$ and $B(4, 5, 8)$ and is perpendicular to the lines $\dfrac{x - 8}{3} = \dfrac{y + 19}{-16} = \dfrac{z - 10}{7}$ and $\dfrac{x - 15}{3} = \dfrac{y - 29}{8} = \dfrac{z - 5}{-5}$.

Concept: Line perpendicular to two given lines — the direction vector of the required line is parallel to the cross product of the direction vectors of the two given lines.

Step 1 – Midpoint of $AB$

Midpoint $M = \left( \frac{2+4}{2}, \frac{3+5}{2}, \frac{4+8}{2} \right) = (3, 4, 6)$.

Step 2 – Direction vectors of given lines

Line 1: $\vec{d_1} = (3, -16, 7)$

Line 2: $\vec{d_2} = (3, 8, -5)$

Step 3 – Direction of required line

$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$

$= \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$ …

We need a line that does two things: it must pass through the midpoint of $A$ and $B$ (since it bisects the segment), and it must be perpendicular to both given lines. A line perpendicular to two lines is parallel to the cross product of their direction vectors — that's the key geometric insight. Once we have a point and a direction, the equation follows directly.

Let's work through it.

1. Find the midpoint of $A$ and $B$. The midpoint $M$ of $A(2,3,4)$ and $B(4,5,8)$ is

This is the point our line must pass through.

2. Extract direction vectors of the given lines. For the line $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$, the direction vector is

For the line $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$, the direction vector is

3. Find a vector perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. The cross product $\vec{d_1} \times \vec{d_2}$ gives a vector perpendicular to both. Compute:

$= (24, 36, 72).$ …