Vector Algebra
Unit Vector and Linear Combination
Grade 12

Question:

<p>Let \(\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\) and \(\vec{b}=\hat{i}+\hat{j}-\hat{k}\). If \(\hat{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\), find \(|\hat{c}\cdot(3\hat{i}-\hat{j}+\hat{k})|\).</p>
<li>\(\dfrac{1}{\sqrt{77}}\)</li>
<li>\(\dfrac{11}{\sqrt{77}}\)</li>
<li>\(\dfrac{7}{\sqrt{77}}\)</li>
<li>\(\sqrt{77}\)</li>

Step-by-Step Solution

Key Concept: c is proportional to a \times b. Compute a \times b, normalise, then take the dot product.
\(\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\1&1&-1\end{vmatrix} =(-2-3)\hat{i}-((-1)-3)\hat{j}+(1-2)\hat{k}=-5\hat{i}+4\hat{j}-\hat{k}\). \(|\vec{a}\times\vec{b}|=\sqrt{25+16+1}=\sqrt{42}\). \(\hat{c}=\dfrac{-5\hat{i}+4\hat{j}-\hat{k}}{\sqrt{42}}\). \(\hat{c}\cdot(3\hat{i}-\hat{j}+\hat{k})=\dfrac{-15-4-1}{\sqrt{42}}=\dfrac{-20}{\sqrt{42}}\). \(|\hat{c}\cdot(3\hat{i}-\hat{j}+\hat{k})|=\dfrac{20}{\sqrt{42}}\). JEE key: C .
Correct Answer: C

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