Question

If the lines \(\dfrac{x-1}{-3}=\dfrac{y-2}{2k}=\dfrac{z-3}{2}\) and \(\dfrac{x-1}{3k}=\dfrac{y-1}{1}=\dfrac{z-6}{-5}\) are mutually perpendicular, find the value of \(k\).

Hint:

Set the dot product of the direction ratios of both lines equal to zero and solve for k.

Answer 1

Step 1: Understanding the Concept:
Two lines in space are perpendicular when their direction vectors have a dot product equal to zero.
We first read off the direction ratios of both lines from their symmetric (Cartesian) form.

Step 2: Identify the Direction Ratios:
Line 1 has direction ratios \((-3,2k,2)\).
Line 2 has direction ratios \((3k,1,-5)\).

Step 3: Apply the Perpendicularity Condition:
For perpendicular lines, \(a_1a_2+b_1b_2+c_1c_2=0\).
\[ (-3)(3k)+(2k)(1)+(2)(-5)=0 \]
Simplify each term and collect like terms.
\[ -9k+2k-10=0 \]
\[ -7k=10 \]

Step 4: Solve for k:
Divide both sides by -7 to isolate k.
\[ k=-\dfrac{10}{7} \]

Final Answer:
The value of k that makes the lines mutually perpendicular is negative ten sevenths. \[ \boxed{k=-\dfrac{10}{7}} \]