Question

Find the angle at which the given lines are inclined to each other: \[ l_1: \frac{x - 5}{2} = \frac{y + 3}{1} = \frac{z - 1}{-3} \] \[ l_2: \frac{x}{3} = \frac{y - 1}{2} = \frac{z + 5}{-1} \]

Hint:

The angle between two lines in space can be calculated using the dot product formula: $\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$, where $\vec{d_1}$ and $\vec{d_2}$ are the direction ratios of the lines.

Answer 1

The direction ratios for line $l_1$ are $\vec{d_1} = \langle 2, 1, -3 \rangle$. The direction ratios for line $l_2$ are $\vec{d_2} = \langle 3, 2, -1 \rangle$. The angle $\theta$ between two lines is calculated using the formula: \[ \cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} \] The dot product is: \[ \vec{d_1} \cdot \vec{d_2} = (2)(3) + (1)(2) + (-3)(-1) = 6 + 2 + 3 = 11 \] The magnitudes of the direction vectors are: \[ |\vec{d_1}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \] \[ |\vec{d_2}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \] Therefore: \[ \cos \theta = \frac{11}{\sqrt{14} \times \sqrt{14}} = \frac{11}{14} \] The angle $\theta$ is: \[ \theta = \cos^{-1} \left( \frac{11}{14} \right) \] Approximating this value gives: \[ \theta \approx 45.57^\circ \]