Q69 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics

If $7 \hat{i}-4 \hat{j}+5 \hat{k}$ is the position vector of the vertex $A$ of a tetrahedron $A B C D$ and $-\hat{i}+4 \hat{j}-3 \hat{k}$ is the position vector of the centroid of the triangle $\mathrm{BCD}$, then the position vector of the centroid of the tetrahedron $A B C D$ is

  • A
    $-\hat{i}+4 \hat{j}-3 \hat{k}$
  • B
    $-\frac{1}{2} \hat{i}+4 \hat{j}-3 \hat{k}$
  • Correct answer
    C
    $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
  • D
    $-\hat{i}-2 \hat{j}+\hat{k}$

Solution

$\begin{aligned} & \text {} \begin{aligned} \bar{A} & =7 \hat{i}-4 \hat{j}+5 \hat{k} \\ \bar{C}_{B C D} & =\frac{\bar{B}+\bar{C}+\bar{D}}{3}=-\hat{i}+4 \hat{j}-3 \hat{k}\end{aligned} \\ & \text { Centroid of tetrahedron } A B C D \\ & \begin{aligned} \bar{C}_{A B C D} & =\frac{1}{4}(\bar{A}+\bar{B}+\bar{C}+\bar{D}) \\ & =\frac{1}{4}\left[\bar{A}+3\left(\frac{\bar{B}+\bar{C}+\bar{D}}{3}\right)\right] \\ & =\frac{1}{4}[7 \hat{i}-4 \hat{j}+5 \hat{k}+3(-\hat{i}+4 \hat{j}-3 \hat{k})] \\ & =\hat{i}+2 \hat{j}-\hat{k} .\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'C'.

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