Tangent of Acute Angle Between Two Lines in 3D Space

To find the angle $\theta$ between the line $L$ and the line $L_3$, we first need to determine the direction vector of line $L$.

The direction vectors of the given lines $L_1$ and $L_2$ are: $\vec{b}_1 = 3\hat{i} + 5\hat{j} + 7\hat{k}$ $\vec{b}_2 = \hat{i} + 4\hat{j} + 7\hat{k}$

Since the line $L$ is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{b}_L$ is parallel to the cross product $\vec{b}_1 \times \vec{b}_2$: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix}$

Evaluating the determinant: $\vec{b}_1 \times \vec{b}_2 = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5) = 7\hat{i} - 14\hat{j} + 7\hat{k}$

Dividing by $7$, we can choose the direction vector of line $L$ as: $\vec{b}_L = \hat{i} - 2\hat{j} + \hat{k}$

The direction vector of line $L_3$ is: $\vec{b}_3 = 2\hat{i} + \hat{j} + 2\hat{k}$

The cosine of the acute angle $\theta$ between $L$ and $L_3$ is given by: $\cos \theta = \frac{|\vec{b}_L \cdot \vec{b}_3|}{|\vec{b}_L| |\vec{b}_3|}$

Calculating the dot product and magnitudes: $\vec{b}_L \cdot \vec{b}_3 = (1)(2) + (-2)(1) + (1)(2) = 2 - 2 + 2 = 2$ $|\vec{b}_L| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$ $|\vec{b}_3| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3$

Substituting these values: $\cos \theta = \frac{2}{3\sqrt{6}}$

Using the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$: $\sec \theta = \frac{3\sqrt{6}}{2}$ $\sec^2 \theta = \frac{9 \times 6}{4} = \frac{54}{4} = \frac{27}{2}$ $\tan^2 \theta = \frac{27}{2} - 1 = \frac{25}{2}$

Since $\theta$ is an acute angle, $\tan \theta > 0$: $\tan \theta = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = \frac{5}{2}\sqrt{2}$