Given, $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $a$ unit vector $\mathbf{c}$ are coplanar. If $\mathbf{c}$ is perpendicular to a, then $\mathbf{c}=$

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Aicharya · Accepted Answer

Given, $\begin{array}{r} \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \ \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{array}$ Let $\mathbf{c}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$ $\begin{array}{lrl} & \therefore & \mathbf{c} \cdot \mathbf{a}=0 \ \Rightarrow & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \ \Rightarrow & 2 x+y+z=0 \quad \ldots (i) \end{array}$ Since, $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are coplanar. $\begin{array}{lc} \therefore & {[\mathbf{a} \mathbf{b} \mathbf{c}]=0} \ \Rightarrow & {\left[\begin{array}{ccc} x & y & z \ 2 & 1 & 1 \ 1 & 2 & -1 \end{array}\right]=0} \ \Rightarrow & x(-1-2)-y(-2-1)+z(4-1)=0 \ \Rightarrow & -3 x+3 y+3 z=0 \ \Rightarrow & -x+y+z=0 \ \Rightarrow & x=y+z \quad \ldots (ii) \end{array}$ By solving Eqs. (i) and (ii), we get $\begin{gathered} x=0, y=-c \ \because \quad x^2+y^2+z^2=1 \Rightarrow y= \pm \frac{1}{\sqrt{2}} \ \therefore \quad \mathbf{c}= \pm \frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{gathered}$

Given, \(\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2…

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