Question

The direction cosines of two lines are connected by the relations \[ l - m + n = 0, \quad 2lm - 3mn + nl = 0. \] If \( \theta \) is the angle between these two lines, then \( \cos \theta \) is:

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{\sqrt{19}} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{3\sqrt{2}} \)

Hint:

For direction cosine problems, use dot product and magnitude formulas to determine the angle.

Answer 1

The Correct Option is B

Let the direction cosines of the two lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
we need to find $\cos \theta = l_1l_2 + m_1m_2 + n_1n_2$ directly.
$l_1 = m_1 - n_1$, $l_2 = m_2 - n_2$.
$2m_1m_2 - 3m_1n_2 - 3m_2n_1 + n_1m_2 + n_2m_1 + n_1n_2 = 0$.
$l = \lambda m - \lambda n$, divide by n, $l/n = t - 1$. $2(t-1)t - 3t + (t-1)=0$
$2t^2-2t-3t+t-1=0$, $2t^2-4t-1=0$.
$t_1,t_2 = \frac{4\pm\sqrt{24}}{4} = 1\pm\frac{\sqrt{6}}{2}$
$l_1l_2+m_1m_2+n_1n_2 = (m_1-n_1)(m_2-n_2)+m_1m_2+n_1n_2=m_1m_2-m_1n_2-m_2n_1+n_1n_2+m_1m_2+n_1n_2=2m_1m_2-m_1n_2-m_2n_1+2n_1n_2$
$2m_1m_2-3n_1m_2+n_1(m_1-n_1)=0$
$2m_1m_2 - n_1n_2 = 2n_1m_2+2n_2m_1-2n_1n_2$
$\cos\theta = \frac{1}{\sqrt{19}}$
Final Answer: The final answer is $\boxed{(2)}$