Question

Let \(\overset{̄}{a},\overset{̄}{b},\overset{̄}{c}\) be unit vectors such that \(\overset{̄}{a}\) is perpendicular to the plane of \(\overset{̄}{b}\) and \(\overset{̄}{c}\). If the angle between \(\overset{̄}{b}\) and \(\overset{̄}{c}\) is \(\frac{π}{3}\), then \(|\overset{̄}{a}+\overset{̄}{b}+\overset{̄}{c}| =\)

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(16\)

Hint:

Square the sum and use the dot products: a is perpendicular to both b and c, and b dot c is cos 60 degrees.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The vector \(\bar{a}\) is perpendicular to the plane that holds \(\bar{b}\) and \(\bar{c}\). So \(\bar{a}\cdot\bar{b} = 0\) and \(\bar{a}\cdot\bar{c} = 0\).

Step 2: Expand the square.
\[ |\bar{a}+\bar{b}+\bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2\bar{a}\cdot\bar{b} + 2\bar{b}\cdot\bar{c} + 2\bar{c}\cdot\bar{a} \]

Step 3: Put the values.
All three are unit vectors, so the squares add to 3. The terms with \(\bar{a}\) vanish. And \(\bar{b}\cdot\bar{c} = \cos\dfrac{\pi}{3} = \dfrac{1}{2}\).
\[ = 3 + 0 + 2\cdot\frac{1}{2} + 0 = 4 \]

Step 4: Take the root.
\(|\bar{a}+\bar{b}+\bar{c}| = \sqrt{4} = 2\).

Final Answer:
The magnitude is 2, option (B). \[ \boxed{2} \]