Question

If \(a,b,c\) are the direction ratios of a line \(L\) and \(\ell,m,n\) are its direction cosines, then \[ \frac{a^2}{b^2+c^2}= \] is equal to:

• A. \(\dfrac{1-\ell^2}{\ell^2}\)
• B. \(\dfrac{\ell^2}{1+\ell^2}\)
• C. \(\dfrac{\ell^2}{\ell^2+m^2}\)
• D. \(\dfrac{\ell^2}{1-\ell^2}\)

Hint:

Always remember: \[ \ell^2+m^2+n^2=1 \] for direction cosines. This identity is frequently used in simplification problems.

Answer 1

The Correct Option is D

Step 1: Relation between direction ratios and direction cosines.
If \[ a,b,c \] are direction ratios of a line, then the corresponding direction cosines are \[ \ell=\frac{a}{\sqrt{a^2+b^2+c^2}}, \] \[ m=\frac{b}{\sqrt{a^2+b^2+c^2}}, \] and \[ n=\frac{c}{\sqrt{a^2+b^2+c^2}} \]

Step 2: Express \(\ell^2\).
Squaring, \[ \ell^2=\frac{a^2}{a^2+b^2+c^2} \] Cross-multiplying, \[ \ell^2(a^2+b^2+c^2)=a^2 \] \[ \ell^2a^2+\ell^2(b^2+c^2)=a^2 \]

Step 3: Rearrange the expression.
Bring the \(a^2\) terms together: \[ \ell^2(b^2+c^2)=a^2-\ell^2a^2 \] \[ \ell^2(b^2+c^2)=a^2(1-\ell^2) \] Now divide both sides by \[ (b^2+c^2)(1-\ell^2) \] \[ \frac{a^2}{b^2+c^2} = \frac{\ell^2}{1-\ell^2} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{a^2}{b^2+c^2}=\frac{\ell^2}{1-\ell^2}} \]