Aspire Faculty ID #18372 · Topic: AMU MCA 2016 · Just now
AMU MCA 2016

If origin is the centroid of a $\triangle PQR$ with vertices $P(2a,2,6)$, $Q(-4,3b,-10)$ and $R(8,14,2c)$, then the value of $a,b,c$ are respectively

Solution

Centroid = $(0,0,0)$ So: $\frac{2a-4+8}{3}=0 \Rightarrow 2a+4=0 \Rightarrow a=-2$ $\frac{2+3b+14}{3}=0 \Rightarrow 3b+16=0 \Rightarrow b=-\frac{16}{3}$ $\frac{6-10+2c}{3}=0 \Rightarrow 2c-4=0 \Rightarrow c=2$

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