Aspire Faculty ID #11960 · Topic: NIMCET 2025 · Just now
NIMCET 2025

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

Solution

Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and the angle between any pair is: 
$\cos^{-1}\left(\frac{1}{4}\right)$ 

Thus, $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=\frac{1}{4}$. 

We are given: $\vec{b}=2\vec{c}+\lambda \vec{a}$ and the magnitude: $|\vec{b}|=4$. 

Since $\vec{b}$ is not a unit vector here, we use: 
$|\vec{b}|^2 = (2\vec{c}+\lambda \vec{a})\cdot(2\vec{c}+\lambda \vec{a})$. 

Expand: $|\vec{b}|^2 = 4|\vec{c}|^2 + \lambda^2|\vec{a}|^2 + 4\lambda(\vec{c}\cdot\vec{a})$ 

Since all are unit vectors: $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$ and $\vec{c}\cdot\vec{a}=\frac14$. 
So: $|\vec{b}|^2 = 4 + \lambda^2 + 4\lambda\left(\frac14\right)$ 
Simplify: $|\vec{b}|^2 = 4 + \lambda^2 + \lambda$ 
Given: $|\vec{b}| = 4 \Rightarrow |\vec{b}|^2 = 16$ 
Thus: $\lambda^2 + \lambda + 4 = 16$ $\lambda^2 + \lambda - 12 = 0$ 
Solve quadratic: $\lambda = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}$ 
So: $\lambda = 3$ (positive root, since $\lambda>0$) 
Final answer: $\boxed{3}$
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