Question

Find the angle between the two lines whose equations are $2x = 3y = -z$ and $6x = -y = -4z$.

• A. 90°
• B. 60°
• C. 30°
• D. 45°

Answer 1

Answer: (A)

90°

Answer 2

The direction vectors of the two lines are found by converting the given equations to their symmetric forms.

For the first line, $2x = 3y = -z$, let this equal $k$. Then $x = k/2$, $y = k/3$, $z = -k$. The direction vector $\vec{a}$ is proportional to $(1/2, 1/3, -1)$. Multiplying by 6 gives $\vec{a} = (3, 2, -6)$.

For the second line, $6x = -y = -4z$, let this equal $m$. Then $x = m/6$, $y = -m$, $z = -m/4$. The direction vector $\vec{b}$ is proportional to $(1/6, -1, -1/4)$. Multiplying by 12 gives $\vec{b} = (2, -12, -3)$.

The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$. The dot product is $\vec{a} \cdot \vec{b} = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$. Since the dot product is 0, the direction vectors are orthogonal, and thus the lines are perpendicular. The angle between them is $90^\circ$.