Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 3, 2020
recategorized
Nov 8, 2020
by Krithiga2101

281 views
0 votes

Given, $a^{2}+b^{2}+c^{2}=1$

We know that, $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

Since square of a number is always positive, then $(a+b+c)^2\geq 0$

$\Rightarrow a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \geq 0$

Now lets examine each option

- $ab+bc+ca =0$

$1+2(0) = 1 \geq 0$ (Possible value)

$\therefore ab+bc+ca =0$ can be a possible value.

- $ab+bc+ca =\cfrac{1}{2}$

$1+2\left(\cfrac{1}{2}\right) = 2 \geq 0$

$\therefore ab+bc+ca =\cfrac{1}{2}$ can be a possible value.

- $ab+bc+ca =\cfrac{-1}{4}$

$1+2\left(\cfrac{-1}{4}\right) = 1-\cfrac{1}{2} = \cfrac{1}{2} \geq 0$

$\therefore ab+bc+ca =\cfrac{-1}{4}$ can be a possible value.

- $ab+bc+ca = -1$

$1+2(-1) = -1 \ngeqslant 0$

$\therefore ab+bc+ca = -1$ can’t be a possible value.

Hence, option D is the correct.

If $a^{2}+b^{2}+c^{2}$ then $ab+bc+ca$ lies in the interval $\left[\cfrac{-1}{2},1\right]$

For proof see this: https://gateoverflow.in/39510/gate2015-ec-2-ga-9

If you know above result then you can directly say answer is option D.