Let the three sides of a triangle ABC be given by the vectors , and . Let G be the centroid of the triangle ABC. Then is equal to __________. [2025]

Question

Let the three sides of a triangle ABC be given by the vectors $2 \hat{i} - \hat{j} + \hat{k}$ , $\hat{i} - 3 \hat{j} - 5 \hat{k}$ and $3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ . Let G be the centroid of the triangle ABC. Then $6 (| \vec{A G} |^{2} + | \vec{B G} |^{2} + | \vec{C G} |^{2})$ is equal to __________. [2025]

Smart Achievers · Accepted Answer

(164)

We have in $△ A B C$ , $\vec{A B} + \vec{C A} = \vec{B C}$

Let PV of $\vec{A}$ be $\vec{0}$ then $\vec{A B} = \vec{B} - \vec{A}$

$\Rightarrow P . V . of \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$

$\vec{C A} = \vec{A} - \vec{C}$

P.V. of $\vec{C} = - \hat{i} + 3 \hat{j} + 5 \hat{k}$

Now, P.V. of $\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{1}{3} (\hat{i} + 2 \hat{j} + 6 \hat{k})$

Then $\vec{A G} = \frac{1}{3} (\hat{i} + 2 \hat{j} + 6 \hat{k})$

$\Rightarrow | \vec{A G} |^{2} = \frac{41}{9}$

$\Rightarrow \vec{B G} = (\frac{1}{3} - 2) \hat{i} + (\frac{2}{3} + 1) \hat{j} + (2 - 1) \hat{k}$

$\Rightarrow | \vec{B G} |^{2} = \frac{59}{9}$

$\Rightarrow \vec{C G} = (\frac{1}{3} + 1) \hat{i} + (\frac{2}{3} - 3) \hat{j} + (2 - 5) \hat{k}$

$\Rightarrow | \vec{C G} |^{2} = \frac{146}{9}$

$∴ 6 (| \vec{A G} |^{2} + | \vec{B G} |^{2} + | \vec{C G} |^{2}) = 6 (\frac{41}{9} + \frac{59}{9} + \frac{146}{9}) = 164$ .

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let the three sides of a triangle ABC be given by the vectors 2i^–j^+k^, i^–3j^–5k^ and 3i^–4j^–4k^. Let G be the centroid of the triangle ABC. Then 6(|AG→|2+|BG→|2+|CG→|2) is equal to __________. [2025]

Want Us to Email you About Special Offers & Updates?