Question

The angle between the lines whose direction cosines are \(\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}\) and \(\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\) is

• A. \(90^{\circ}\)
• B. \(120^{\circ}\)
• C. \(45^{\circ}\)
• D. \(30^{\circ}\)

Hint:

Use cos of the angle equals the sum of products of the corresponding direction cosines.

Answer 1

The Correct Option is B

Step 1: Check the direction cosines
For both lines, $\dfrac{3}{16}+\dfrac{1}{16}+\dfrac{3}{4}=1$, so they are valid direction cosines.

Step 2: Dot product
Only the sign of the third component differs, so the dot product is $\dfrac{4}{16}-\dfrac{12}{16}=-\dfrac{1}{2}$.

Step 3: Result
The angle whose cosine is $-\dfrac{1}{2}$ is $120^{\circ}$.

Final Answer:
The angle is 120 degrees. \[ \boxed{\text{(B) }120^{\circ}} \]