Let $\overrightarrow{\mathrm{OA}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}},…

Let $\overrightarrow{\mathrm{OA}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{OC}}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ be the position vectors of three points, $\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively. If $\mathrm{G}$ is the centroid of triangle $\mathrm{ABC}$, then $\mathrm{BC}^2+\mathrm{CA}^2+\mathrm{AB}^2+9(\mathrm{OG})^2=$
  1. 162
  2. 156
  3. 144
  4. 132

Solution

$ \begin{aligned} & \text { Position vector of centroid }=\overrightarrow{\mathrm{OG}}=\frac{\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}}{3} \\ & \Rightarrow \overrightarrow{\mathrm{OG}}=-2^{\wedge} \end{aligned} $ Now $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=-4 \hat{i}+5 \hat{j}+5 \hat{k}$ $ \begin{aligned} & \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}=-5 \hat{i}-4 \hat{j}-\hat{k} \\ & \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=-\hat{i}-\hat{j}-4 \hat{k} \end{aligned} $ Therefore $ \begin{aligned} & \mathrm{BC}^2+\mathrm{CA}^2+\mathrm{AB}^2+9(\mathrm{OG})^2=\overrightarrow{\mathrm{BC}} \cdot \overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AB}} \\ & +9(\overrightarrow{\mathrm{OG}} \cdot \overrightarrow{\mathrm{OG}}) \end{aligned} $ $\begin{aligned} & =66+42+18+9 \\ & =162\end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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