Question

If \(\overset{̄}{a},\overset{̄}{b}\) and \(\overset{̄}{c}\) are non-coplanar unit vectors such that the angle between any two of them is \(60^{\circ}\), and the vector \(\overset{̄}{d} = x\overset{̄}{a}+y\overset{̄}{b}+z\overset{̄}{c}\) is perpendicular to both \(\overset{̄}{a}\) and \(\overset{̄}{b}\), then the value of \(\frac{(x+y)}{z}\) is

• A. \(-\frac{1}{2}\)
• B. \(-\frac{2}{3}\)
• C. \(-1\)
• D. \(0\)

Hint:

Dot \(\vec d\) with \(\vec a\) and \(\vec b\); each dot product between different unit vectors is \(\frac12\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The vectors are unit vectors at \(60^{\circ}\), so \(\vec a\cdot\vec a = 1\) and \(\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = \frac12\).

Step 2: Use perpendicularity:
\(\vec d\cdot\vec a = x + \frac{y}{2} + \frac{z}{2} = 0\).
\(\vec d\cdot\vec b = \frac{x}{2} + y + \frac{z}{2} = 0\).

Step 3: Solve:
Subtract: \(\frac{x}{2} - \frac{y}{2} = 0\), so \(x = y\).
Put \(y = x\) in the first equation: \(x + \frac{x}{2} + \frac{z}{2} = 0\), so \(3x + z = 0\) and \(z = -3x\).
\[ \frac{x+y}{z} = \frac{2x}{-3x} = -\frac23 \]

Final Answer:
The value is \(-\frac23\), option (B). \[ \boxed{-\frac{2}{3}} \]