Question:
If \(\vec x=3\hat i-6\hat j-\hat k,\ \vec y=\hat i+4\hat j-3\hat k\) and \(\vec z=3\hat i-4\hat j-12\hat k\), then the magnitude of the projection of \(\vec x\times\vec y\) on \(\vec z\) is :
If \(\vec x=3\hat i-6\hat j-\hat k,\ \vec y=\hat i+4\hat j-3\hat k\) and \(\vec z=3\hat i-4\hat j-12\hat k\), then the magnitude of the projection of \(\vec x\times\vec y\) on \(\vec z\) is :
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Compute \(\vec x\times\vec y\), dot with \(\vec z\), and divide by \(|\vec z|=13\).
Updated On: Nov 11, 2026
- \(12\)
- \(15\)
- \(14\)
- \(13\)
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The Correct Option is C
Solution and Explanation
\(\vec x\times\vec y=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-6&-1\\1&4&-3\end{vmatrix}=(18+4)\hat i-(-9+1)\hat j+(12+6)\hat k=22\hat i+8\hat j+18\hat k\).
\(|\vec z|=\sqrt{9+16+144}=13\).
\((\vec x\times\vec y)\cdot\vec z=66-32-216=-182\).
Projection \(=\frac{-182}{13}=-14\); magnitude \(=14\).
Answer: (3) \(14\)
\(|\vec z|=\sqrt{9+16+144}=13\).
\((\vec x\times\vec y)\cdot\vec z=66-32-216=-182\).
Projection \(=\frac{-182}{13}=-14\); magnitude \(=14\).
Answer: (3) \(14\)
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