Question

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-\frac{8}{7}}{2} = \frac{y-\frac{4}{7}}{1} = \frac{z}{2}$, then $\tan \theta$ is equal to:

• A. $\frac{3\sqrt{2}}{2}$
• B. $\frac{5\sqrt{2}}{2}$
• C. $\frac{5\sqrt{2}}{3}$
• D. $\frac{4\sqrt{2}}{3}$

Hint:

The direction of a line perpendicular to two other lines can be found using the cross product of their direction vectors. Then use the formula for the angle between two lines.

Answer 1

The Correct Option is B

We are given two lines $L_1$ and $L_2$ and a third line $L$ perpendicular to both. This implies that the direction of $L$ is along the normal to the plane containing directions of $L_1$ and $L_2$.

Let the direction ratios of $L$ be $(l, m, n)$. Since $L \perp L_1$ and $L \perp L_2$:
1) $3l + 5m + 7n = 0$
2) $1l + 4m + 7n = 0$
Subtracting (2) from (1): $2l + m = 0 \implies m = -2l$.
Substituting $m = -2l$ into (2): $l + 4(-2l) + 7n = 0 \implies l - 8l + 7n = 0 \implies -7l + 7n = 0 \implies n = l$.
So the ratios $l:m:n$ are $l : -2l : l$, which simplify to $1 : -2 : 1$.

The line $L_3$ has direction ratios $(2, 1, 2)$.
The angle $\theta$ between $L$ and $L_3$ is given by:
$$\cos \theta = \frac{|1 \cdot 2 + (-2) \cdot 1 + 1 \cdot 2|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}} = \frac{2}{\sqrt{6} \cdot 3}$$
We know that $\sec^2 \theta = 1 + \tan^2 \theta$.
$$\sec \theta = \frac{1}{\cos \theta} = \frac{3\sqrt{6}}{2}$$
$$\sec^2 \theta = \frac{9 \times 6}{4} = \frac{54}{4} = \frac{27}{2}$$
$$\tan^2 \theta = \sec^2 \theta - 1 = \frac{27}{2} - 1 = \frac{25}{2}$$
$$\tan \theta = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$$