Question

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a} = \vec{b} \times (\vec{b} \times \vec{c})$. If magnitudes of the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are $\sqrt{2}, 1$ and 2 respectively and the angle between $\vec{b}$ and $\vec{c}$ is $\theta$ ($0<\theta<\frac{\pi}{2}$), then the value of $1+\tan\theta$ is equal to :

• A. 1
• B. 2
• C. $\sqrt{3}+1$
• D. $\frac{\sqrt{3}+1}{\sqrt{3}}$

Hint:

The vector triple product identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$ is extremely useful in simplifying expressions involving nested cross products.

Answer 1

The Correct Option is B

Given: \[ \vec{a} = \vec{b} \times (\vec{b} \times \vec{c}) \] Using the vector triple product identity, \[ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A}\cdot\vec{C})\vec{B} - (\vec{A}\cdot\vec{B})\vec{C} \] we get: \[ \vec{a} = (\vec{b}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{b})\vec{c} \] Now, \[ \vec{b}\cdot\vec{c} = |\vec{b}||\vec{c}|\cos\theta = (1)(2)\cos\theta = 2\cos\theta \] \[ \vec{b}\cdot\vec{b} = |\vec{b}|^2 = 1 \] Hence, \[ \vec{a} = 2\cos\theta\,\vec{b} - \vec{c} \] Taking magnitude squared on both sides: \[ |\vec{a}|^2 = |2\cos\theta\,\vec{b} - \vec{c}|^2 \] \[ |\vec{a}|^2 = 4\cos^2\theta|\vec{b}|^2 + |\vec{c}|^2 - 2(2\cos\theta)(\vec{b}\cdot\vec{c}) \] Substituting values: \[ 2 = 4\cos^2\theta + 4 - 8\cos^2\theta \] \[ 2 = 4 - 4\cos^2\theta \Rightarrow \cos^2\theta = \frac{1}{2} \] Since \( 0 < \theta < \frac{\pi}{2} \), \[ \cos\theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{\pi}{4} \] \[ \tan\theta = 1 \Rightarrow 1 + \tan\theta = 2 \] \[ \boxed{2} \]