If $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4…

If $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}},-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are respectively the position vectors of four coplanar points $P, Q, R$ and $S$, then $\lambda=$
  1. $\frac{46}{17}$
  2. $-\frac{46}{17}$
  3. $\frac{146}{17}$
  4. $-\frac{146}{17}$

Solution

Given, $ \begin{aligned} & \mathbf{P}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{Q}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \mathbf{R}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $ $\begin{aligned} & \text { and } \quad \mathbf{S}=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} \\ & \text { Here, } \quad \text { PS }=(4-3) \hat{\mathbf{i}}+(5+2) \hat{\mathbf{j}}+(\lambda+1) \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+(\lambda+1) \hat{\mathbf{k}} \\ & \mathbf{P Q}=-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ & \text { and } \mathbf{P R}=-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \end{aligned}$ Now, $ \begin{aligned} \mathbf{P Q} \times & \mathbf{P R}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 5 & -3 \\ -4 & 3 & 3 \end{array}\right| \\ & =\hat{\mathbf{i}}(15+9)-\hat{\mathbf{j}}(-3-12)+\hat{\mathbf{k}}(-3+20) \\ & =24 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+17 \hat{\mathbf{k}} \end{aligned} $ Since, $P, Q, R$ and $S$ are coplanar, $ \begin{array}{lc} \therefore & \mathbf{P S} \cdot(\mathbf{P Q} \times \mathbf{P R})=0 \\ \Rightarrow & {[\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+(\lambda+1) \hat{\mathbf{k}}] \cdot[24 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+17 \hat{\mathbf{k}}]=0} \\ \Rightarrow & 24+105+17(\lambda+1)=0 \Rightarrow 129+17 \lambda+17=0 \\ \Rightarrow & 17 \lambda=-146 \quad \Rightarrow \lambda=-\frac{146}{17} \end{array} $

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

Practice more Vectors questions on Aicharya