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A unit vector perpendicular to the plane containing the vectors \( \hat{i}+2 \hat{j}+\hat{k} \) and
\( -2 \hat{i}+\hat{j}+3 \hat{k} \) is
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\( -2 \hat{i}+\hat{j}+3 \hat{k} \) is
Solution:
1217 Upvotes
Verified Answer
The correct answer is:
\(-\hat{i}+\hat{j}-\hat{k} \)
(D)
\[
\begin{array}{l}
\hat{n}=\left(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right) \\
\vec{a} \times \vec{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
-2 & 1 & 3
\end{array}\right|=5 \hat{i}-5 \hat{j}+5 \hat{k} \\
|\vec{a} \times \vec{b}|=5 \sqrt{3} \\
\therefore \hat{n}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}} \\
\text { or } \hat{n}=\frac{-\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}
\end{array}
\]
\[
\begin{array}{l}
\hat{n}=\left(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right) \\
\vec{a} \times \vec{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
-2 & 1 & 3
\end{array}\right|=5 \hat{i}-5 \hat{j}+5 \hat{k} \\
|\vec{a} \times \vec{b}|=5 \sqrt{3} \\
\therefore \hat{n}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}} \\
\text { or } \hat{n}=\frac{-\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}
\end{array}
\]
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