Question

The volume of the tetrahedron formed $4\hat{i}+5\hat{j}+\hat{k},-\hat{j}+\hat{k},3\hat{i}+9\hat{j}+4\hat{k},4(-\hat{i}+\hat{j}+\hat{k})$ is:

• A. $7$
• B. $9$
• C. $11$
• D. $13$

Answer 1

The correct option is C $11$

$\overrightarrow{a}=4\hat{i}+5\hat{j}+\hat{k}$

$\overrightarrow{b}=(-1)\hat{j}+\hat{k}$

$\overrightarrow{c}=3\hat{i}+9\hat{j}+4\hat{k}$

$\overrightarrow{d}=(-4)\hat{i}+4\hat{j}+4\hat{k}$

Let us take the tetrahedral to be originating from the origin.

ie, the three vectors that makes up a tetrahedron are, $(\overrightarrow{a}-\overrightarrow{d}),(\overrightarrow{b}-\overrightarrow{d}),(\overrightarrow{c}-\overrightarrow{d})$

$\therefore$ Volume of the Tetrahedron

$=\frac{\left[\right(\overrightarrow{a}-\overrightarrow{d}),(\overrightarrow{b}-\overrightarrow{d}),(\overrightarrow{c}-\overrightarrow{d}\left)\right]}{6}=\frac{\left|\begin{array}{ccc}(4-(-4\left)\right)& (5-4)& (1-4)\\ (0-(-4\left)\right)& \left(\right(-1)-4)& (1-4)\\ (3-(-4\left)\right)& (9-4)& (4-4)\end{array}\right|}{6}=\frac{\left|\begin{array}{ccc}8& 1& (-3)\\ 4& (-5)& (-3)\\ 7& 5& 0\end{array}\right|}{6}=\frac{\left|7\right((-3)-\left(15\right))-5((-24)+12\left)\right|}{6}=|5\times 2-7\times 3|=11$