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Question

Mathematics Question on Plane

The volume of the tetrahedron with vertices P (-1, 2, 0), Q ( 2, 1, -3), R (1, 0, 1) and S (3, -2, 3) is

A

13\frac{1}{3}

B

23\frac{2}{3}

C

14\frac{1}{4}

D

34\frac{3}{4}

Answer

23\frac{2}{3}

Explanation

Solution

Given : The vertices of tetrahedron are
P(-1, 2, 0), Q(2, 1, -3), R(1, 0, 1) & S(3, -2, 3)
∴\therefore Volume of tetrahedron = 16 [PQ⃗ PR⃗ PS⃗]= \ \frac{1}{6} \ [\vec{PQ} \ \vec{PR} \ \vec{PS}]
Now,
PQ⃗=(2+1)i^+(1−2)j^+(−3)k^=3i^−j^−3k^\vec{PQ}=(2+1)\hat{i}+(1-2)\hat{j}+(-3)\hat{k}=3\hat{i}-\hat{j}-3\hat{k}
Similarly, PR⃗=2i^−2j^+k^\vec{PR}=2\hat{i}-2\hat{j}+\hat{k}
&     PS⃗=4i^−4j^+3k^\& \ \ \ \ \ \vec{PS}=4\hat{i}-4\hat{j}+3\hat{k}
∴\therefore Volume of tetrahedron

=16∣3−1−3 2−21 4−43∣=23=\frac{1}{6} \begin{vmatrix} 3 & -1 & -3\\\ 2 & -2 & 1 \\\ 4 & -4 & 3 \end{vmatrix} =\frac{2}{3}