The direction ratios of the line perpendicular to the lines \(\displaystyle\frac{x-7}{-6}=\frac{y+17}{4}=\frac{z-6}{2}\) and \(\displaystyle\frac{x+5}{6}=\frac{y+3}{3}=\frac{z-4}{-6}\) are proportional to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5, 4, 7
The direction ratios of the two lines give the vectors
\(\vec{b}_1 = -6\hat{\imath} + 4\hat{\jmath} + 2\hat{k}\)
\(\vec{b}_2 = 6\hat{\imath} + 3\hat{\jmath} - 6\hat{k}\)
A line perpendicular to both is parallel to their cross product:
\( \displaystyle \vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ -6 & 4 & 2\\ 6 & 3 & -6\end{vmatrix} \)
\(= \hat{\imath}(-24 - 6) - \hat{\jmath}(36 - 12) + \hat{k}(-18 - 24)\)
\(= -30\hat{\imath} - 24\hat{\jmath} - 42\hat{k}\)
Dividing by −6, the direction ratios are proportional to 5, 4, 7.
Shortcut: The required direction ratios must give a zero dot product with both lines. For 5, 4, 7:
\(-6(5) + 4(4) + 2(7) = -30 + 16 + 14 = 0\)
\(6(5) + 3(4) - 6(7) = 30 + 12 - 42 = 0\)
The first check already fails for the other options: 7, 4, 5 gives −16; 7, 5, 4 gives −14; 5, 7, 4 gives 6.