Question

If \(\overset{̄}{a}\), \(\overset{̄}{b}\) and \(\overset{̄}{c}\) are three vectors such that \(|\overset{̄}{a}+\overset{̄}{b}+\overset{̄}{c}| = 1\), \(\overset{̄}{c} = λ(\overset{̄}{a}\times \overset{̄}{b})\) and \(|\overset{̄}{a}| = \frac{1}{\sqrt{3}}\), \(|\overset{̄}{b}| = \frac{1}{\sqrt{2}}\), \(|\overset{̄}{c}| = \frac{1}{\sqrt{6}}\), then the angle between \(\overset{̄}{a}\) and \(\overset{̄}{b}\) is

• A. \(\frac{π^c}{6}\)
• B. \(\frac{π^c}{4}\)
• C. \(\frac{π^c}{3}\)
• D. \(\frac{π^c}{2}\)

Hint:

c is perpendicular to a and b, so |a+b+c| squared splits as |a+b| squared plus |c| squared.

Answer 1

The Correct Option is D

Step 1: Check the magnitudes:
$|\bar a|^2 + |\bar b|^2 + |\bar c|^2 = \frac13 + \frac12 + \frac16 = 1$. This is exactly $|\bar a + \bar b + \bar c|^2$.

Step 2: Interpretation:
For three vectors, $|\bar a + \bar b + \bar c|^2 = |\bar a|^2 + |\bar b|^2 + |\bar c|^2 + 2(\bar a\cdot\bar b + \bar b\cdot\bar c + \bar c\cdot\bar a)$. The dot products involving $\bar c$ are 0, so $\bar a\cdot\bar b = 0$.

Step 3: Angle:
$\cos\theta = 0$, so $\theta = \frac\pi2$.

Final Answer:
The angle is pi by 2, option (D). \[ \boxed{\frac{\pi}{2}} \]