Question

Let \(\overset{⃗}{a}\) and \(\overset{⃗}{b}\) be linearly independent vectors such that
\(|\overset{⃗}{a}| = \sqrt{3},|\overset{⃗}{b}| = 3\) and \(|\overset{⃗}{a}-\overset{⃗}{b}| = 4\).
If \(\overset{⃗}{a}\times (2\hat{i}+2\hat{j}-\hat{k}) = (2\hat{i}+2\hat{j}-\hat{k})\times \overset{⃗}{b}\) and \(|(\overset{⃗}{a}+\overset{⃗}{b})\cdot (3\hat{i}+4\hat{j}+2\hat{k})| = \sqrt{λ}\), then \(λ = \ldots\)

• A. \(32\)
• B. \(64\)
• C. \(256\)
• D. \(128\)

Hint:

Use a + b + c = 0 to get b.c from the magnitudes.

Answer 1

The Correct Option is D

Step 1: Use the sum condition:
\(\vec a = -(\vec b + \vec c)\), so \(|\vec a|^2 = |\vec b|^2 + |\vec c|^2 + 2\vec b\cdot\vec c\).

Step 2: Find cos theta:
\(9 = 9 + 9 + 2\cdot9\cos\theta\), so \(\cos\theta = -\frac12\) and \(\theta = 120^\circ\).

Step 3: Evaluate:
\(\tan^2\theta = (\sqrt3)^2 = 3\) and \(\cot^2\theta = \frac13\). So \(\tan^2\theta + \cot^2\theta = 3 + \frac13 = \frac{10}{3}\).

Final Answer:
The value is \(\frac{10}{3}\), option (D). \[ \boxed{\frac{10}{3}} \]