If the line \(\displaystyle\frac{-x+1}{3}=\frac{-y-2}{-2k}=\frac{z+3}{2}\) and \(\displaystyle\frac{-1+x}{3k}=\frac{-1+y}{1}=\frac{-z+6}{5}\) are perpendicular, then the value of k is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{10}{7}\)
Direction ratios are read only after writing each line in the standard form \(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}\), with coefficient +1 for x, y and z.
First line:
\( \displaystyle \frac{x-1}{-3} = \frac{y+2}{2k} = \frac{z+3}{2} \)
\(\text{direction ratios } (-3,\ 2k,\ 2)\)
Second line:
\( \displaystyle \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5} \)
\(\text{direction ratios } (3k,\ 1,\ -5)\)
The lines are perpendicular when \(a_1a_2 + b_1b_2 + c_1c_2 = 0\):
\((-3)(3k) + (2k)(1) + (2)(-5) = 0\)
\(-9k + 2k - 10 = 0\)
\(-7k = 10\)
\( \displaystyle k = -\frac{10}{7} \)
Tip: In \(\frac{-x+1}{3}\) the direction ratio for x is −3, not 3. Always make the coefficients of x, y, z equal to +1 before reading the denominators.