Question

Volume of the parallelepiped having vertices at $O\equiv (0,0,0),A\equiv (2,-2,1),B\equiv (5,-4,4)$ and $C\equiv (1,-2,4)$ is

• A. $5$ cu. units
• B. $10$ cu. units
• C. $15$ cu. units
• D. $20$ cu. units

Answer 1

The correct option is B

$10$ cu. units

Explanation for the correct option

Given: $O\equiv (0,0,0),A\equiv (2,-2,1),B\equiv (5,-4,4)$ and $C\equiv (1,-2,4)$

$\begin{array}{rcl}OA\left[(0,0,0),(2,-2,1)\right]& =& (2-0)i+(-2-0)j+(1-0)k\\ & =& 2i-2j+k\end{array}$

$\begin{array}{rcl}OB\left[(0,0,0),(5,-4,4)\right]& =& \left(5-0\right)i+(-4-0)j+(4-0)k\\ & =& 5i-4j+4k\end{array}$

$\begin{array}{rcl}OC\left[(0,0,0),(1,-2,4)\right]& =& (1-0)i+(-2-0)j+(4-0)k\\ & =& i-2j+4k\end{array}$

Therefore volume of a parallelepiped$=l\times b\times h$

$$\begin{gathered} =OA\times OB\times OC \\ =(2i-2j+k)\times (5i-4j+4k)\times (i-2j+4k) \\ =\left|\begin{array}{ccc}2& -2& 1\\ 5& -4& 4\\ 1& -2& 4\end{array}\right| \end{gathered}$$

We know that $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=a_1(b_2{c}_3-b_3{c}_2)-a_2(b_1{c}_3-b_3{c}_1)+a_3(b_1{c}_2-b_2{c}_1)$

$\begin{array}{rcl}& \Rightarrow & \left|\begin{array}{ccc}2& -2& 1\\ 5& -4& 4\\ 1& -2& 4\end{array}\right|=2\left[-4\times 4-\left(4\right)\times (-2)\right]-(-2)\left[5\times 4-4\times 1\right]+\left(1\right)\left[5\times (-2)-(-4)\times 1\right]\\ & =& 2(-16+8)+2(20-4)+(-10+4)\\ & =& 2(-8)+2\left(16\right)-6\\ & =& -16+32-6\\ & =& 10cuunits\end{array}$

Hence, option B is correct.