🎥 Video solution / Text Solution of this question is given below:
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:
$\boxed{3}$