Question

The angle between the lines \(\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}\) and \(\frac{x+1}{\mathrm{~m}}=\frac{y-3}{\mathrm{n}}=\frac{\mathrm{z}-1}{l}\), where \(l>\mathrm{m}>\mathrm{n}\) and \(1, \mathrm{~m}, \mathrm{n}\) are roots of the equation \(x^3+x^2-4 x-4=0\), is

• A. \(\mathrm{cos}^{-1}(\frac{2}{9})\)
• B. \(\mathrm{cos}^{-1}(\frac{-4}{9})\)
• C. \(\mathrm{cos}^{-1}(\frac{2}{3})\)
• D. \(\mathrm{cos}^{-1}(\frac{1}{9})\)

Hint:

Factorise the cubic to find l, m, n in order, then use the dot product of the direction ratios.

Answer 1

The Correct Option is B

Step 1: Roots by trial
$x=-1$ satisfies the cubic: $-1+1+4-4=0$. The quotient is $x^2-4$, giving $x=\pm2$.

Step 2: Magnitudes
Both direction vectors have magnitude $\sqrt{4+1+4}=3$.

Step 3: Result
The dot product is $-4$, so $\cos\theta=-\dfrac{4}{9}$.

Final Answer:
The angle is cos inverse of -4/9. \[ \boxed{\text{(B) }\cos^{-1}\left(\dfrac{-4}{9}\right)} \]