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

To find the direction of line $L$, we recognize that it is perpendicular to two given lines $L_1$ and $L_2$. The direction ratios (D.R.s) of $L_1$ are $(3, 5, 7)$ and those of $L_2$ are $(1, 4, 7)$.

The direction vector $\vec{d}$ of line $L$ will be proportional to the cross product of the direction vectors of $L_1$ and $L_2$:
$$\vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$$
Calculating the determinant:
$$\vec{d} = \mathbf{i}(5 \times 7 - 4 \times 7) - \mathbf{j}(3 \times 7 - 1 \times 7) + \mathbf{k}(3 \times 4 - 1 \times 5)$$
$$\vec{d} = \mathbf{i}(35 - 28) - \mathbf{j}(21 - 7) + \mathbf{k}(12 - 5) = 7\mathbf{i} - 14\mathbf{j} + 7\mathbf{k}$$
Dividing by 7, we get simpler D.R.s for $L$: $(1, -2, 1)$.

Now, the direction ratios of line $L_3$ are $(2, 1, 2)$.
Let $\theta$ be the angle between $L$ and $L_3$. Using the cosine formula for the angle between two lines with D.R.s $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:
$$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$$
$$\cos \theta = \frac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}} = \frac{|2 - 2 + 2|}{\sqrt{6} \cdot \sqrt{9}} = \frac{2}{3\sqrt{6}}$$
To find $\tan \theta$, we first find $\sin \theta$:
$$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{4}{54} = 1 - \frac{2}{27} = \frac{25}{27}$$
$$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{25/27}{2/27} = \frac{25}{2}$$
$$\tan \theta = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$$