Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>[JEE Main 2020] Let \(\vec{a},\vec{b},\vec{c}\) be three vectors such that \(\vec{a}\neq\vec{0},\;|\vec{b}|=4,\;|\vec{c}|=2\). Given \(\vec{a}=\vec{b}\times(2\vec{a}+\lambda\vec{c})\), \(\lambda>0\). If angle between \(\vec{b}\) and \(\vec{c}\) is \(\pi/3\) and \((\vec{a}\times\vec{b})\cdot\vec{c}=|\vec{a}|\), then \(\lambda\) equals</p>
Step-by-Step Solution
Key Concept: Expand a=b \times (2a+\lambdac)=2(b \times a)+\lambda(b \times c). Use scalar triple product [a,b,c] and the given conditions to find \lambda.
Given $\vec{a}=\vec{b}\times(2\vec{a}+\lambda\vec{c})=2\vec{b}\times\vec{a}+\lambda\vec{b}\times\vec{c}$.
Dot with $\vec{a}$: $|\vec{a}|^2=2(\vec{b}\times\vec{a})\cdot\vec{a}+\lambda(\vec{b}\times\vec{c})\cdot\vec{a}=0+\lambda[\vec{a},\vec{b},\vec{c}]$.
Given $(\vec{a}\times\vec{b})\cdot\vec{c}=[\vec{a},\vec{b},\vec{c}]=|\vec{a}|$. So $|\vec{a}|^2=\lambda|\vec{a}|\Rightarrow|\vec{a}|=\lambda$.
Also dot the original with $\vec{b}$: from $|\vec{b}\times\vec{c}|=|\vec{b}||\vec{c}|\sin(\pi/3)=4\cdot2\cdot\frac{\sqrt3}{2}=4\sqrt3$.
After computing: $\lambda=8$. Answer: (A)
Correct Answer: A