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: Understanding the Concept:
Because \(\bar c = \lambda(\bar a \times \bar b)\), the vector \(\bar c\) is perpendicular to both \(\bar a\) and \(\bar b\), hence perpendicular to \(\bar a + \bar b\).

Step 2: Use Pythagoras:
\[ |\bar a + \bar b + \bar c|^2 = |\bar a + \bar b|^2 + |\bar c|^2 \]
So \(1 = |\bar a + \bar b|^2 + \frac16\), giving \(|\bar a + \bar b|^2 = \frac56\).

Step 3: Expand:
\[ |\bar a + \bar b|^2 = |\bar a|^2 + |\bar b|^2 + 2\bar a\cdot\bar b = \frac13 + \frac12 + 2\bar a\cdot\bar b = \frac56 + 2\bar a\cdot\bar b \]
Equating with \(\frac56\) gives \(\bar a\cdot\bar b = 0\).

Step 4: Conclusion:
The dot product is zero, so the angle is \(\frac{\pi}{2}\).

Step 5: Why the other options are wrong.
Angles \(\frac\pi6, \frac\pi4, \frac\pi3\) would give a non-zero dot product \(|\bar a||\bar b|\cos\theta\), which would make \(|\bar a + \bar b|^2\) differ from \(\frac56\).

Final Answer:
The angle is \(\frac\pi2\), option (D). \[ \boxed{\frac{\pi}{2}} \]