Question

If the direction cosines of a line are \[ \left( \frac{a}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{c}{\sqrt{83}} \right) \] and \(c-a=4\), then \(ca=\)

• A. \(24\)
• B. \(21\)
• C. \(18\)
• D. \(33\)

Hint:

For direction cosines \((l,m,n)\), always remember: \[ l^2+m^2+n^2=1 \] This identity is very useful for finding unknown quantities in 3D geometry problems.

Answer 1

The Correct Option is B

Step 1: Use the property of direction cosines.
If \[ (l,m,n) \] are the direction cosines of a line, then \[ l^2+m^2+n^2=1 \] Here, \[ l=\frac{a}{\sqrt{83}}, \qquad m=\frac{5}{\sqrt{83}}, \qquad n=\frac{c}{\sqrt{83}} \] Therefore, \[ \left(\frac{a}{\sqrt{83}}\right)^2+ \left(\frac{5}{\sqrt{83}}\right)^2+ \left(\frac{c}{\sqrt{83}}\right)^2=1 \] \[ \frac{a^2}{83}+\frac{25}{83}+\frac{c^2}{83}=1 \] Multiplying throughout by \(83\), \[ a^2+25+c^2=83 \] \[ a^2+c^2=58 \]

Step 2: Use the given condition.
Given, \[ c-a=4 \] Squaring both sides, \[ (c-a)^2=16 \] \[ c^2+a^2-2ac=16 \] Using \[ a^2+c^2=58, \] we get \[ 58-2ac=16 \] \[ 2ac=42 \] \[ ac=21 \]

Step 3: Final conclusion.
Therefore, \[ \boxed{21} \]