If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be 3 non-coplanar vectors and $(\mathbf{a}-\lambda \mathbf{b})…

If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be 3 non-coplanar vectors and $(\mathbf{a}-\lambda \mathbf{b}),(\mathbf{b}-2 \mathbf{c}) \times(\mathbf{c}+2 \mathbf{a})=0$, then $\lambda$ is equal to
  1. 1
  2. $\frac{1}{4}$
  3. 0
  4. $\frac{-1}{4}$

Solution

a, b, c are there non-coplanar vectors $ \begin{aligned} & (\mathbf{a}-\lambda \mathbf{b}) \cdot[(\mathbf{b}-2 \mathbf{c}) \times(\mathbf{c}+2 \mathbf{a})]=0 \\ & \Rightarrow(\mathbf{a}-\lambda \mathbf{b}) \cdot\{\mathbf{b} \times \mathbf{c}+2(\mathbf{b} \times \mathbf{a})-2(\mathbf{c} \times \mathbf{c}) \\ & -4(\mathbf{c} \times \mathbf{a})\}=0 \\ & \Rightarrow(\mathbf{a}-\lambda \mathbf{b}) \cdot \mathbf{b} \times \mathbf{c}+2(\mathbf{b} \times \mathbf{a})-4(\mathbf{c} \times \mathbf{a})\}=0 \\ & {[\because \mathbf{a} \times \mathbf{a}=\mathbf{0}]} \\ & \Rightarrow \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})+2 \mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})-4 \mathbf{a} \cdot(\mathbf{c} \times \mathbf{a}) \\ & -\lambda \mathbf{b} \cdot(\mathbf{b} \times \mathbf{c})-2 \lambda \mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})+4 \lambda \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})=0 \\ & \Rightarrow[\mathbf{a} \mathbf{b} \mathbf{c}]+2[\mathbf{a} \mathbf{b} \mathbf{a}]-4[\mathbf{a} \mathbf{c} \mathbf{a}]-\lambda[\mathbf{b} \mathbf{b} \mathbf{c}] \\ & -2 \lambda[\mathbf{b} \mathbf{b} \mathbf{a}]+4 \lambda[\mathbf{b} \mathbf{c} \mathbf{a}]=0 \end{aligned} $ $\therefore$ We know that in scalar triple product, if two vectors are same in box then its value is 0 . $\begin{aligned} & \Rightarrow[\mathbf{a} \mathbf{b} \mathbf{c}]+2(0)-4(0)-\lambda(0)+2 \lambda(0)+4 \lambda[\mathbf{b} \mathbf{c} \mathbf{a}]=0 \\ & \Rightarrow[\mathbf{a} \mathbf{b} \mathbf{c}]=-4 \lambda[\mathbf{b} \mathbf{c} \mathbf{a}] \\ & \Rightarrow \quad 1=-4 \lambda \quad\left[\because[\mathbf{a} \mathbf{b} \mathbf{c}]=\left[\begin{array}{ll}\mathbf{b} & \mathbf{c} \mathbf{a}]]\end{array}\right.\right. \\ & \Rightarrow \quad \lambda=-\frac{1}{4} \\ & \end{aligned}$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

Practice more Vectors questions on Aicharya