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: Write the fundamental identity for direction cosines.
If $ l,m,n $ are direction cosines: $ l^2+m^2+n^2=1 $. This is the defining Pythagorean property in 3D.
Step 2: Isolate $ m^2+n^2 $.
\[ m^2+n^2 = 1-l^2 \]
Step 3: Recall the proportionality between direction ratios and cosines.
$ b = \frac{am}{l} $ and $ c = \frac{an}{l} $ (since $ a:b:c = l:m:n $).
Step 4: Express $ b^2+c^2 $ in terms of $ a,l,m,n $.
\[ b^2+c^2 = \frac{a^2m^2}{l^2}+\frac{a^2n^2}{l^2} = \frac{a^2(m^2+n^2)}{l^2} \]
Step 5: Form the required ratio.
\[ \frac{a^2}{b^2+c^2} = \frac{a^2}{\frac{a^2(m^2+n^2)}{l^2}} = \frac{l^2}{m^2+n^2} \]
Step 6: Substitute $ m^2+n^2 = 1-l^2 $.
\[ \frac{a^2}{b^2+c^2} = \frac{l^2}{1-l^2} \] \[ \boxed{\dfrac{l^2}{1-l^2}} \]