Question

The altitude of a parallelopiped whose three coterminous edges are the vectors,$\overline{A}=\hat{i}+\hat{j}+\hat{k};\overline{B}=2\hat{i}+4\hat{j}-\hat{k}$ & $\overline{C}=\overline{i}+\overline{j}+3\hat{k}$ with $\overline{A}$ and $\overline{B}$ as the sides of the base of the parallelopiped is

• A. $2/\sqrt{19}$
• B. $4/\sqrt{19}$
• C. $2\sqrt{38}/19$
• D. none

Answer 1

Answer: (C)

$2\sqrt{38}/19$

$\overrightarrow{A}\times \overrightarrow{B}=(\hat{i}+\hat{j}+\hat{k})\times (2\hat{i}+4\hat{j}-\hat{k})=-5\hat{i}+3\hat{j}+2\hat{k}$

Area of base $S=|\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{(-5{)}^2+3^2+2^2}=\sqrt{38}$

Volume of parallelepiped, $V=\overrightarrow{C}.(\overrightarrow{A}\times \overrightarrow{B})$

$\therefore$ $V=(\hat{i}+\hat{j}+3\hat{k}).(-5\hat{i}+3\hat{j}+2\hat{k})=4$

$⟹$ Altitude $h=\frac{V}{S}=\frac{4}{\sqrt{38}}=\frac{2\sqrt{38}}{19}$