What will be the value of $$r^{2}+p^{2}+s^{2}-rp-ps-rs$$ where $$r = 1, p = 2, s = 3$$
Given, $$r = 1, p = 2, s = 3$$
$$\therefore\ r^2+p^2+s^2-rp-ps-rs=\left(1\right)^2+\left(2\right)^2+\left(3\right)^2-\left(1\right)\left(2\right)-\left(2\right)\left(3\right)-\left(1\right)\left(3\right)=1+4+9-2-6-3=3$$
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