Question

If $x,y$ and $z$ are three unit vectors in three-dimensional space, then the minimum value of $\mid \hat{x}+\hat{y}{\mid }^2+\mid \hat{y}+\hat{z}{\mid }^2+\mid \hat{z}+\hat{x}{\mid }^2$ is :

• A. $\frac{3}{2}$
• B. $3$
• C. $3\sqrt{3}$
• D. $6$

Answer 1

The correct option is B $3$

We have ${|\hat{x}+\hat{y}|}^2$

$=(\hat{x}+\hat{y}).\widehat{(x}+\hat{y})$

$=|\hat{x}{|}^2+|\hat{y}{|}^2+2\hat{x}.\hat{y}$

$=2+2\hat{x}.\hat{y}$

Similarly ,

${|\hat{y}+\hat{z}|}^2=2+2\hat{y}.\hat{z}$

and ${|\hat{x}+\hat{z}|}^2=2+2\hat{z}.\hat{x}$

Given that
$\Rightarrow {|\hat{x}+\hat{y}|}^2+{|\hat{y}+\hat{z}|}^2+{|\hat{x}+\hat{z}|}^2=6+2(\hat{x}.\hat{y}+\hat{y}.\hat{z}+\hat{z}.\hat{x})$.....$\left(1\right)$

Now, ${|\hat{x}+\hat{y}+\hat{z}|}^2={\left|\hat{x}\right|}^2+{\left|\hat{y}\right|}^2+{\left|\hat{z}\right|}^2+2(\hat{x}.\hat{y}+\hat{y}.\hat{z}+\hat{z}.\hat{x})\Rightarrow {|\hat{x}+\hat{y}+\hat{z}|}^2=3+2(\hat{x}.\hat{y}+\hat{y}.\hat{z}+\hat{z}.\hat{x})$

Hence, ${|\hat{x}+\hat{y}+\hat{z}|}^2\ge 0\Rightarrow 3+2(\hat{x}.\hat{y}+\hat{y}.\hat{z}+\hat{z}.\hat{x})\ge 0\Rightarrow 2(\hat{x}.\hat{y}+\hat{y}.\hat{z}+\hat{z}.\hat{x})\ge -3$

Using the above result in equation $\left(1\right)$, we get the minumum value of the expression in question as $3$