Question

The angle between the lines $\dfrac{x-3}{-4} = \dfrac{y+2}{3} = \dfrac{z-1}{5}$ and $\dfrac{x-2}{2} = \dfrac{y-4}{1} = \dfrac{z+3}{3}$ is

• A. $\cos^{-1}\!\left(\dfrac{1}{3\sqrt{7}}\right)$
• B. $\cos^{-1}\!\left(\dfrac{\sqrt{2}}{1}\right)$
• C. $\cos^{-1}\!\left(\dfrac{2}{7}\right)$
• D. $\cos^{-1}\!\left(\dfrac{1}{2\sqrt{7}}\right)$
• E. $\cos^{-1}\!\left(\dfrac{1}{\sqrt{7}}\right)$

Hint:

Always take the absolute value of the dot product before dividing when computing the acute angle between two lines. This ensures $\theta \in [0°, 90°]$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The angle $\theta$ between two lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ satisfies $\cos\theta = \dfrac{|\vec{d_1}\cdot\vec{d_2}|}{|\vec{d_1}||\vec{d_2}|}$.

Step 2: Detailed Explanation:
$\vec{d_1} = (-4,3,5)$, $\vec{d_2} = (2,1,3)$.
$\vec{d_1}\cdot\vec{d_2} = (-4)(2)+(3)(1)+(5)(3) = -8+3+15 = 10$.
$|\vec{d_1}| = \sqrt{16+9+25} = \sqrt{50} = 5\sqrt{2}$.
$|\vec{d_2}| = \sqrt{4+1+9} = \sqrt{14}$.
\[ \cos\theta = \frac{|10|}{5\sqrt{2}\cdot\sqrt{14}} = \frac{10}{5\sqrt{28}} = \frac{10}{10\sqrt{7}} = \frac{1}{\sqrt{7}} \] \[ \theta = \cos^{-1}\!\left(\frac{1}{\sqrt{7}}\right) \]

Step 3: Final Answer:
The angle is $\cos^{-1}\!\left(\dfrac{1}{\sqrt{7}}\right)$.