Question

Let $\vec{a}$ and $\vec{c}$ be unit vectors such that the angle between them is $\cos^{-1} \left( \frac{1}{4} \right)$. If $\vec{b} = 2\vec{c} + \lambda \vec{a}$. Where $\lambda > 0$ and $|\vec{b}| = 4$, then $\lambda$ is equal to: 
 

• A. 4
• B. 3
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Hint:

When dealing with vector magnitudes and angles, break down the problem into dot product components and use the Pythagorean theorem to solve for unknowns.

Answer 1

The Correct Option is B

We are given that $\vec{a}$ and $\vec{c}$ are unit vectors, so $|\vec{a}| = 1$ and $|\vec{c}| = 1$. The angle between $\vec{a}$ and $\vec{c}$ is $\cos^{-1} \left( \frac{1}{4} \right)$. Using the cosine rule: \[ \vec{a} \cdot \vec{c} = \cos^{-1} \left( \frac{1}{4} \right) = \frac{1}{4} \] We are also given that $|\vec{b}| = 4$ and $\vec{b} = 2\vec{c} + \lambda \vec{a}$. The magnitude of $\vec{b}$ is given by: \[ |\vec{b}|^2 = (2\vec{c} + \lambda \vec{a}) \cdot (2\vec{c} + \lambda \vec{a}) \] Expanding: \[ |\vec{b}|^2 = 4|\vec{c}|^2 + 4\lambda (\vec{c} \cdot \vec{a}) + \lambda^2 |\vec{a}|^2 \] Since $|\vec{a}|^2 = 1$ and $|\vec{c}|^2 = 1$, and $\vec{c} \cdot \vec{a} = \frac{1}{4}$, we get: \[ 16 = 4 + 4\lambda \cdot \frac{1}{4} + \lambda^2 \] Simplifying: \[ 16 = 4 + \lambda + \lambda^2 \] \[ \lambda^2 + \lambda - 12 = 0 \] Solving the quadratic equation: \[ \lambda = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm \sqrt{49}}{2} = \frac{-1 \pm 7}{2} \] Thus, $\lambda = 3$ (since $\lambda > 0$).