Q78 / 180 · COMEDK 2023 (Memory Based) · Mathematics

If the vectors $\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{c}=m \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are coplanar, then the value of $m$ is

  • A
    $\frac{5}{8}$
  • Correct answer
    B
    $\frac{8}{5}$
  • C
    $\frac{-7}{4}$
  • D
    $\frac{2}{3}$

Solution

Since, vectors a, b and c are coplanar $\therefore[\mathbf{a} b \mathbf{c}]=0 \Rightarrow \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0$ Now, $\begin{aligned} \mathbf{b} \times \mathbf{c} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ m & -1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(4-1)-\hat{\mathbf{j}}(2+m)+\hat{\mathbf{k}}(-1-2 m) \\ & =3 \hat{\mathbf{i}}-(2+m) \hat{\mathbf{j}}-(1+2 m) \hat{\mathbf{k}} \end{aligned}$ $\begin{aligned} \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) & =(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}-(2+m) \hat{\mathbf{j}}-(1+2 m) \hat{\mathbf{k}}) \\ & =2(3)+3(2+m)-4(1+2 m) \\ & =6+6+3 m-4-8 m=8-5 m \\ & \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\ & 8-5 m=0 \Rightarrow m=\frac{8}{5} \end{aligned}$ and $\begin{aligned} \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) & =(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}-(2+m) \hat{\mathbf{j}}-(1+2 m) \hat{\mathbf{k}}) \\ & =2(3)+3(2+m)-4(1+2 m) \\ & =6+6+3 m-4-8 m=8-5 m \\ & \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\ & 8-5 m=0 \Rightarrow m=\frac{8}{5} \end{aligned}$

Hence, the correct answer is option: 'B'.

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