Question

If the two diagonals of one of the faces of a parallelopiped are $6\hat{i}+6\hat{k}$ and $4\hat{j}+2\hat{k}$ and one of the edges not containing the given diagonals is $4\hat{j}-8\hat{k}$, then the volume of the parallelopiped is

• A. $60$
• B. $80$
• C. $100$
• D. $120$

Answer 1

The correct option is D $120$

Let $\overrightarrow{a}=6\hat{i}+6\hat{k},\overrightarrow{b}=4\hat{j}+2\hat{k}$ and $\overrightarrow{c}=4\hat{j}-8\hat{k}$
Then $\overrightarrow{a}\times \overrightarrow{b}=-24\hat{i}-12\hat{j}+24\hat{k}$
$=12(-2\hat{i}-\hat{j}+2\hat{k})$
$\therefore$ Area of the base of parallelopiped $=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$
$=\frac{1}{2}(12\times 3)=18$
Height of parallelopiped $=$ Length of projection of $\overrightarrow{c}$ on $\overrightarrow{a}\times \overrightarrow{b}$
$=\frac{|\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}|}{|\overrightarrow{a}\times \overrightarrow{b}|}$
$=\frac{12(-4-16)}{36}=\frac{20}{3}$

Volume of parallelopiped $=18\times \frac{20}{3}=120$