If the origin is the centroid of the triangle whose vertices are A(2,p,−3),B(q,−2,5) and R(−5,1,r) , then find the values of p,q,r.

Engineering

Mathematics

theorem in space

Section Formulae and Centres of a Triangle

Distance from a Plane

Question

If the origin is the centroid of the triangle whose vertices are A(2,p,−3),B(q,−2,5) and R(−5,1,r) , then find the values of p,q,r.

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Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of ΔABC whose vertices are

A(2,p,– 3), B(q,–2, 5) R(–5,1,r)

∴ \vec{a} = 2 \hat{i} + p \hat{j} - 3 \hat{k}, \vec{b} = q \hat{i} - 2 \hat{j} + 5 \hat{k}, \vec{c} = - 5 \hat{i} + \hat{j} + r \hat{k}

Given that O is the centroid of the triangle ABC

∴ (0, 0, 0) = \frac{\vec{a} + \vec{b} + \vec{c}}{3}

\Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0}

\Rightarrow 2 \hat{i} + p \hat{j} - 3 \hat{k} + q \hat{i} - 2 \hat{j} + 5 \hat{k} - 5 \hat{i} + \hat{j} + r \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}

\Rightarrow (2 + q - 5) \hat{i} + (p - 2 + 1) \hat{j} + (- 3 + 5 + r) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}

By equating the corresponding terms we get

2 + q − 5 = 0 ⇒ q = 3

p − 2 + 1 = 0 ⇒ p = 1

− 3 + 5 + r = 0 ⇒ r = −2

∴ p = 1, q = 3, r = −2