The value of \(\lambda\) for which the angle between the vectors \(\hat{\imath}+\lambda\hat{\jmath}+\hat{k}\) and \(\hat{\imath}+\hat{\jmath}\) is \(\displaystyle\frac{\pi}{3}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Let \(\vec{a} = \hat{\imath}+\lambda\hat{\jmath}+\hat{k}\) and \(\vec{b} = \hat{\imath}+\hat{\jmath}\). Using \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\):
\(\vec{a}\cdot\vec{b} = 1+\lambda\)
\(|\vec{a}| = \sqrt{2+\lambda^2}, \quad |\vec{b}| = \sqrt{2}\)
\( \displaystyle \frac{1+\lambda}{\sqrt{2}\,\sqrt{2+\lambda^2}} = \frac{1}{2} \)
\(2(1+\lambda) = \sqrt{2(2+\lambda^2)}\)
The left side must be positive, so \(\lambda \gt -1\). Squaring:
\(4+8\lambda+4\lambda^2 = 4+2\lambda^2\)
\(2\lambda(\lambda+4) = 0\)
\(\lambda = 0 \text{ or } \lambda = -4\)
Only \(\lambda = 0\) satisfies \(\lambda \gt -1\).
Check: for \(\lambda = 0\), \(\cos\theta = \displaystyle\frac{1}{\sqrt{2}\times\sqrt{2}} = \frac{1}{2}\).
Why the other options are incorrect
Option (1) −4 — Incorrect. It comes from squaring, but then \(\cos\theta = \displaystyle\frac{-3}{\sqrt{2}\times\sqrt{18}} = -\frac{1}{2}\), so the angle is \(\displaystyle\frac{2\pi}{3}\), not \(\displaystyle\frac{\pi}{3}\).
Option (2) 4 — Incorrect. Then \(\cos\theta = \displaystyle\frac{5}{\sqrt{2}\times\sqrt{18}} = \frac{5}{6}\), not \(\displaystyle\frac{1}{2}\).
Option (4) 2 — Incorrect. Then \(\cos\theta = \displaystyle\frac{3}{\sqrt{2}\times\sqrt{6}} = \frac{\sqrt{3}}{2}\), so the angle is \(\displaystyle\frac{\pi}{6}\).