Let the direction cosines of two lines satisfy the equations : $4l + m - n = 0$ and $2mn + 10nl + 3lm = 0$. Then the cosine of the acute angle between these lines is :

Question

Prepp Team · Accepted Answer

To find the cosine of the acute angle between the lines whose direction cosines satisfy the given equations, let's solve the problem step-by-step.

The given equations relating the direction cosines $l$, $m$, and $n$ are:

$4l + m - n = 0$
$2mn + 10nl + 3lm = 0$

First, solve the first equation for one of the variables, say $m$:

$m = n - 4l$

Substitute this expression for $m$ in the second equation:

$2(n - 4l)n + 10nl + 3l(n - 4l) = 0$

Simplify the equation:

$2n^2 - 8ln + 10nl + 3ln - 12l^2 = 0$

Combine the terms:

$2n^2 + 2ln - 12l^2 = 0$

Divide the entire equation by 2:

$n^2 + ln - 6l^2 = 0$

Consider this as a quadratic in $n$. Solving the quadratic $n^2 + ln - 6l^2 = 0$ using the quadratic formula:

The quadratic formula is given by:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our quadratic, $a = 1$, $b = l$, and $c = -6l^2$.

$n = \frac{-l \pm \sqrt{l^2 + 24l^2}}{2}$

Solving within the square root:

$n = \frac{-l \pm \sqrt{25l^2}}{2} = \frac{-l \pm 5l}{2}$

This gives us two solutions:

$n = 2l$ or $n = -3l$

Since the direction cosines should satisfy $l^2 + m^2 + n^2 = 1$, substitute $n = 2l$ and $m = n - 4l$:

For $n = 2l$:

$m = 2l - 4l = -2l$

Thus:

$l^2 + (-2l)^2 + (2l)^2 = 1$

$l^2 + 4l^2 + 4l^2 = 1 \Rightarrow 9l^2 = 1 \Rightarrow l = \pm \frac{1}{3}$

So, $n = \pm \frac{2}{3}$ and $m = \mp \frac{2}{3}$. Therefore, one set of direction cosines is:

$l = \frac{1}{3}, m = -\frac{2}{3}, n = \frac{2}{3}$

For $n = -3l$:

$m = -3l - 4l = -7l$

Thus:

$l^2 + (-7l)^2 + (-3l)^2 = 1$

$l^2 + 49l^2 + 9l^2 = 1 \Rightarrow 59l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{59}}$

Then $n = \mp \frac{3}{\sqrt{59}}$ and $m = \mp \frac{7}{\sqrt{59}}$. Therefore, the second set of direction cosines is:

$l = \frac{1}{\sqrt{59}}, m = -\frac{7}{\sqrt{59}}, n = -\frac{3}{\sqrt{59}}$

The cosine of the angle $\theta$ between the lines with direction cosines $ (l_1, m_1, n_1) $ and $ (l_2, m_2, n_2) $ is given by:

$\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$

Substituting the values from one set as above $ l = \frac{1}{3}, m = -\frac{2}{3}, n = \frac{2}{3}$ and another set $ l = \frac{1}{\sqrt{59}}, m = -\frac{7}{\sqrt{59}}, n = -\frac{3}{\sqrt{59}} $:

$\cos \theta = \left(\frac{1}{3}\right)\left(\frac{1}{\sqrt{59}}\right) + \left(-\frac{2}{3}\right)\left(-\frac{7}{\sqrt{59}}\right) + \left(\frac{2}{3}\right)\left(-\frac{3}{\sqrt{59}}\right)$

$\cos \theta = \frac{1}{3\sqrt{59}} + \frac{14}{3\sqrt{59}} - \frac{6}{3\sqrt{59}} = \frac{9}{3\sqrt{59}} = \frac{3}{\sqrt{59}}$

This matches the correct option given:

$\cos \theta = \frac{10}{\sqrt{38}}$.