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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
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Verified Answer
The correct answer is:
$\frac{8}{5}$
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
$\begin{aligned}
\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) & =(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}-(2+m) \hat{\mathbf{j}}-(1+2 m) \hat{\mathbf{k}}) \\
& =2(3)+3(2+m)-4(1+2 m) \\
& =6+6+3 m-4-8 m=8-5 m \\
& \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\
& 8-5 m=0 \Rightarrow m=\frac{8}{5}
\end{aligned}$
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
$\begin{aligned}
\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) & =(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}-(2+m) \hat{\mathbf{j}}-(1+2 m) \hat{\mathbf{k}}) \\
& =2(3)+3(2+m)-4(1+2 m) \\
& =6+6+3 m-4-8 m=8-5 m \\
& \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\
& 8-5 m=0 \Rightarrow m=\frac{8}{5}
\end{aligned}$
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