Question

The volume of a tetrahedron whose vertices are
$4\hat{i}+5\hat{j}+\hat{k},-\hat{j}+\hat{k},3\hat{i}+9\hat{j}+4\hat{k}$ and
$-2\hat{i}+4\hat{j}+4\hat{k}$ is (in cubic units)

• A. $\frac{14}{3}$
• B. 5
• C. 6
• D. 30

Answer 1

Answer: (B)

Given vertices of a tetrahedron are
$A(4\hat{i}+5\hat{j}+\hat{k}),B(-\hat{j}+\hat{k})$
$C(3\hat{i}+9j+4k),D(-2\hat{i}+4\hat{j}+4\hat{k})$
So, $AB=-4\hat{i}-6\hat{j}AC=-\hat{i}+4\hat{j}+3\hat{k}$
$AD=-6\hat{i}-\hat{j}+3\hat{k}$
Now, volume of the given tetrahedron is
$v=\frac{1}{6}|\left|\begin{array}{ccc}<br/><br/>-4& -6& 0\\ <br/><br/>-1& 4& 3\\ <br/><br/>-6& -1& 3<br/><br/>\end{array}\right|\mid$
$=\frac{1}{6}\times |[-4(12+3)+6(-3+18)+0]|$
$=\frac{1}{6}|[-60+90]|=\frac{30}{6}=5$ cubic units