Question

The direction cosines of a line which is perpendicular to the lines \(\frac{x-7}{2} = \frac{y+17}{-3} = \frac{z-6}{1}\) and \(\frac{x+5}{1} = \frac{y+3}{2} = \frac{z-6}{-2}\) are...

• A. \(\frac{\pm 4}{3\sqrt{10}},\frac{\mp 5}{3\sqrt{10}},\frac{\pm 7}{3\sqrt{10}}\)
• B. \(\frac{\pm 4}{3\sqrt{10}},\frac{\pm 5}{3\sqrt{10}},\frac{\mp 7}{3\sqrt{10}}\)
• C. \(\frac{\mp 4}{3\sqrt{10}},\frac{\pm 5}{3\sqrt{10}},\frac{\pm 7}{3\sqrt{10}}\)
• D. \(\frac{\pm 4}{3\sqrt{10}},\frac{\pm 5}{3\sqrt{10}},\frac{\pm 7}{3\sqrt{10}}\)

Hint:

Take the cross product of the direction vectors.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A line perpendicular to two lines is parallel to the cross product of their direction vectors.

Step 2: Direction vectors:
\(\vec d_1=(2,-3,1)\) and \(\vec d_2=(1,2,-2)\).

Step 3: Cross product:
\[ \vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-3&1\\1&2&-2\end{vmatrix}=\hat i(6-2)-\hat j(-4-1)+\hat k(4+3)=4\hat i+5\hat j+7\hat k \]

Step 4: Magnitude:
\(\sqrt{16+25+49}=\sqrt{90}=3\sqrt{10}\).

Step 5: Direction cosines:
\[ \pm\frac{4}{3\sqrt{10}},\ \pm\frac{5}{3\sqrt{10}},\ \pm\frac{7}{3\sqrt{10}} \]
The signs are all together plus or all together minus. Option (D). Options (A), (B) and (C) change relative signs, so their ratios are not 4:5:7.

Final Answer:
The direction cosines are all plus or all minus 4, 5, 7 over 3 sqrt 10. \[ \boxed{\pm\frac{4}{3\sqrt{10}},\pm\frac{5}{3\sqrt{10}},\pm\frac{7}{3\sqrt{10}}} \]