Vector Algebra
Scalar Triple Product
Grade None
Question:
<p>The altitude of a parallelepiped whose three coterminous edges are \(\vec{A}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{B}=2\hat{i}+4\hat{j}-\hat{k}\), \(\vec{C}=\hat{i}+\hat{j}+3\hat{k}\), with \(\vec{A}\) and \(\vec{B}\) as the base, is</p>
\(2\sqrt{19}\)
\(\dfrac{4}{\sqrt{19}}\)
\(\dfrac{2\sqrt{38}}{19}\)
none of these
Step-by-Step Solution
Key Concept: Volume = |[A,B,C]|. Base area = |A \times B|. Altitude h = Volume/Base area.
Volume: \([A,B,C]=\begin{vmatrix}1&1&1\\2&4&-1\\1&1&3\end{vmatrix}\)
\(=1(12+1)-1(6+1)+1(2-4)=13-7-2=4\).
Base area (A\timesB): \(\vec{A}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\2&4&-1\end{vmatrix}=(-1-4)\hat{i}-(-1-2)\hat{j}+(4-2)\hat{k}=-5\hat{i}+3\hat{j}+2\hat{k}\).
\(|\vec{A}\times\vec{B}|=\sqrt{25+9+4}=\sqrt{38}\).
Altitude \(h=\dfrac{4}{\sqrt{38}}=\dfrac{4\sqrt{38}}{38}=\dfrac{2\sqrt{38}}{19}\). Answer: (C)
Correct Answer: C