If the volume of tetrahedron, whose vertices are $\mathrm{A}(1,2,3), \mathrm{B}(-3,-1,1), \mathrm{C}(2,1,3)$…

If the volume of tetrahedron, whose vertices are $\mathrm{A}(1,2,3), \mathrm{B}(-3,-1,1), \mathrm{C}(2,1,3)$ and $\mathrm{D}(-1,2, x)$ is $\frac{11}{6}$ cubic units, then the value of $x$ is
  1. $3$
  2. $-2$
  3. $4$
  4. $-1$

Solution

$\begin{aligned} & \overline{\mathrm{AB}}=-4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \overline{\mathrm{AC}}=\hat{\mathrm{i}}-\hat{\mathrm{j}} \\ & \overline{\mathrm{AD}}=-2 \hat{\mathrm{i}}+(x-3) \hat{\mathrm{k}} \\ & \text { Volume }=\frac{1}{6}\left|\begin{array}{ccc}-4 & -3 & -2 \\ 1 & -1 & 0 \\ -2 & 0 & x-3\end{array}\right| \\ \therefore \quad & 11=4(x-3)+3(x-3)-2(-2) \\ \therefore \quad & 11=7 x-17\end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 2)

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