Q.Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.

Concept: Perpendicular Vectors Condition — a line perpendicular to two given lines has its direction vector parallel to the cross product of the direction vectors of those lines.

Step 1: Direction vectors of the given lines:

$\vec{d_1} = (3, -16, 7)$, $\vec{d_2} = (3, 8, -5)$.

Step 2: Direction vector of the 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 the vector equation of a line that passes through a fixed point and is perpendicular to two given lines. The key idea: a line perpendicular to two lines must be parallel to the cross product of their direction vectors. Why? Because if a line is perpendicular to each of two lines, its direction vector must be perpendicular to both direction vectors — and the cross product gives exactly that: a vector orthogonal to both.

Let’s extract the direction vectors from the given symmetric equations.

1. First line: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$

   Its direction vector is $\vec{d}_1 = (3, -16, 7)$.

2. Second line: $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$

   Its direction vector is $\vec{d}_2 = (3, 8, -5)$.

3. Find a vector perpendicular to both: compute the cross product $\vec{d}_1 \times \vec{d}_2$.

Expand:

• $\hat{i}$ component: $(-16)(-5) - (7)(8) = 80 - 56 = 24$
• $\hat{j}$ component: $-( (3)(-5) - (7)(3) ) = -(-15 - 21) = -(-36) = 36$
• $\hat{k}$ component: $(3)(8) - (-16)(3) = 24 + 48 = 72$

So $\vec{d}_1 \times \vec{d}_2 = (24, 36, 72)$.

4. Simplify the direction: any scalar multiple works. Divide by 12: $(2, 3, 6)$. This is a cleaner direction vector for our required line. …

Method: Direction of a Line Perpendicular to Two Given Lines

Use this when you must build a line perpendicular to two given lines and passing through a stated point — the cross product hands you the direction in one step.

Steps

Step 1: Extract the two direction vectors.

From each given line take its direction ratios, $\vec{b_1}$ and $\vec{b_2}$. The points on the given lines are not used for the direction.

Step 2: Take the cross product for the required direction.

$\vec{b}=\vec{b_1}\times\vec{b_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}$ …

Common Mistakes

Mistake 1: Using a point from one of the two given lines as the fixed point.

Why it's wrong: the required line must pass through the specified point $(1,2,-4)$; the given lines only supply directions, not the point. Correct approach: use $(1,2,-4)$ as $\vec{a}$ and the cross product as the direction.

Mistake 2: Trying to guess a perpendicular direction instead of taking the cross product. …