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The unit vectors orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and coplanar with \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) are
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Verified Answer
The correct answer is:
\(\pm \frac{1}{\sqrt{10}}(3 \hat{\mathbf{j}}-\hat{\mathbf{k}})\)
A vector which is coplanar to the vectors \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is
\(\mathbf{a}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathbf{b}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\)
\((2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}\) ...(i)
Eq. (i) is orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\)
\(\begin{aligned}
& 3(2 a+b)+2(a-b)+6(a+b) =0 \\
& 14 a+7 b =0 \\
& 2 a+b =0
\end{aligned}\)
Required unit vector,
\(\begin{aligned}
& = \pm \frac{(2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}}{\sqrt{(2 a+b)^2+(a-b)^2+(a+b)^2}} \\
& = \pm \frac{3 a \hat{\mathbf{j}}-a \hat{\mathbf{k}}}{\sqrt{(3 a)^2+a^2}}= \pm \frac{3 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{10}}
\end{aligned}\)
Hence, option (b) is correct.
\(\mathbf{a}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathbf{b}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\)
\((2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}\) ...(i)
Eq. (i) is orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\)
\(\begin{aligned}
& 3(2 a+b)+2(a-b)+6(a+b) =0 \\
& 14 a+7 b =0 \\
& 2 a+b =0
\end{aligned}\)
Required unit vector,
\(\begin{aligned}
& = \pm \frac{(2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}}{\sqrt{(2 a+b)^2+(a-b)^2+(a+b)^2}} \\
& = \pm \frac{3 a \hat{\mathbf{j}}-a \hat{\mathbf{k}}}{\sqrt{(3 a)^2+a^2}}= \pm \frac{3 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{10}}
\end{aligned}\)
Hence, option (b) is correct.
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