Exam

JEE Main

Year

2025

Date

4 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Vector Algebra

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Vector Algebra

Let the three sides of a triangle A B C be given by the vectors 2 i...

Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \quad \hat{i}-3 \hat{j}-5 \hat{k} and$3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let$G$be the centroid of the triangle$A B C$. Then$6$\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$ is equal to ________

Step-by-step solution

By given data
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CB}}$
Let pv of $\overrightarrow{\mathrm{A}}$ are $\overrightarrow{\mathrm{O}}$ then
$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}$
i.e. p v of $\overrightarrow{\mathrm{B}}-=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{C}}$
i.e. pv of $\overrightarrow{\mathrm{C}}=-(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
Now pv of centroid
$$\begin{aligned}<br/>& (\overrightarrow{\mathrm{G}})-=\frac{\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}}{3}=\frac{\overrightarrow{0}+(2,-1,1)+(-1,3,5)}{3} \\ & \overrightarrow{\mathrm{G}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})<br/>\end{aligned}$$
Now $\overrightarrow{\mathrm{AG}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$
$\Rightarrow|\overrightarrow{\mathrm{AG}}|^2=\frac{1}{9} \times 41$
$\overrightarrow{\mathrm{BG}}=\left(\frac{1}{3}-2\right) \hat{\mathrm{i}}+\left(\frac{2}{3}+1\right) \hat{\mathrm{j}}+(2-1) \hat{\mathrm{k}}$
$\Rightarrow|\overrightarrow{\mathrm{BG}}|^2=\frac{59}{9}$
$$\begin{aligned}<br/>& \overrightarrow{\mathrm{CG}}=\left(\frac{1}{3}+1\right) \hat{\mathrm{i}}+\left(\frac{2}{3}-3\right) \hat{\mathrm{j}}+(2-5) \hat{\mathrm{k}} \\ & \Rightarrow|\overrightarrow{\mathrm{CG}}|^2=\frac{146}{9}<br/>\end{aligned}$$
Now
$$\begin{aligned}<br/>& 6\left[|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right]=6 \times\left[\frac{41}{9}+\frac{59}{9}+\frac{146}{9}\right] \\ & =6 \times \frac{246}{9}=164<br/>\end{aligned}$$