How Good Are You In Vectors?

$\begin{aligned} \left|\vec b \times \vec c \right| & = 2 \\ \implies \left| \vec b \right| \left|\vec c \right|\color{#3D99F6} \sin \theta & = 2 & \small \color{#3D99F6} \text{where }\vec b \cdot \vec c = \cos \theta. \\ 4(1)\sin \theta & = 2 \\ \implies \theta & = \frac \pi 6 \end{aligned}$

$\begin{aligned} \therefore \vec b & = 4\left(\cos \frac \pi 6{\color{#3D99F6}\vec c} + \sin \frac \pi 6 {\color{#3D99F6} \vec d}\right) & \small \color{#3D99F6} \text{where }\vec c \text{ and } \vec d \text{ are two} \\ & = 4\left(\frac {\sqrt 3}2{\color{#3D99F6} \vec c} + \frac 12{\color{#3D99F6} \vec d} \right) & \small \color{#3D99F6} \text{orthogonal unit vectors.} \\ 2 \vec b & = 4\sqrt 3 \vec c + 4\vec d \\ & = \vec c + \color{#3D99F6} (4\sqrt 3 - 1)\vec c + 4\vec d & \small \color{#3D99F6} \text{Given: }2\vec b = \vec c + \lambda \vec a \\ \implies \lambda \vec a & = (4\sqrt 3 - 1)\vec c + 4\vec d \\ \left|\lambda \vec a \right| & = \left|(4\sqrt 3 - 1)\vec c + 4\vec d\right| \\ \implies \lambda & = \sqrt{(4\sqrt 3 - 1)^2 + 4^2} \\ & = \boxed{\sqrt{65-8\sqrt 3}} \end{aligned}$

Using vector cross product |b x c| = |b|.|c|.sin (b,c), we find the angle between vector b, and c, 30 deg.Using cos rule for the triangle formed by vectors c,2b, and and λ(a,) ⃗ (| λa|)^2 = |2b|^2 + |c|^2 – 2|2b||c| cos θ. λ^2 = 64 + 1 - 8√3 , Then we find λ = √(65+8√3)