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: Solve with unknowns:
Let the line have ratios $(l,m,n)$. Perpendicular to the first: $2l-3m+n=0$. To the second: $l+2m-2n=0$.

Step 2: Eliminate n:
From the first, $n=3m-2l$. Substitute: $l+2m-6m+4l=0$, so $5l=4m$.

Step 3: Choose m = 5:
Then $l=4$ and $n=15-8=7$. Ratios $4:5:7$.

Step 4: Normalise:
Divide by $\sqrt{90}=3\sqrt{10}$. Option (D).

Final Answer:
The ratios 4 : 5 : 7 normalise to option D. \[ \boxed{D} \]