Q.If the lines $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular, find the value of $k$.

Concept: Perpendicular Vectors Condition — two lines are perpendicular when the dot product of their direction vectors is zero.

Step 1: Direction vector of first line:

$\vec{d_1} = (-3, 2k, 2)$

Step 2: Direction vector of second line:

$\vec{d_2} = (3k, 1, -5)$

Step 3: Perpendicular condition: $\vec{d_1} \cdot \vec{d_2} = 0$ …

The direction vectors of the two lines are

Perpendicular lines require $\vec b_1\cdot\vec b_2 = 0$: …

Method: Finding an Unknown Parameter from a Perpendicularity Condition

Use this when two lines contain an unknown (like $k$) in their direction ratios and you are told the lines are perpendicular; the condition turns into a single equation for the unknown.

Steps

Step 1: Extract each direction vector, keeping the unknown symbolic.

From the symmetric form $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$, the denominators are the direction ratios. Write both as $\vec{b_1}$ and $\vec{b_2}$ with the unknown left in place.

Step 2: Impose perpendicularity as dot product $=0$.

This is the one condition perpendicular lines must satisfy. …

Common Mistakes

Mistake 1: Using the proportionality (parallel) condition instead of the dot-product (perpendicular) condition.

Why it's wrong: for perpendicular lines you need $\vec{b_1}\cdot\vec{b_2}=0$, not $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$. Setting up proportions here gives a wrong equation for $k$. Correct approach: because the lines are perpendicular, write $(-3)(3k)+(2k)(1)+(2)(-5)=0$. …