Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be non-coplanar vectors and $\overrightarrow{d}$ be a non-zero vector which is perpendicular to $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$. If $\overrightarrow{d}=x\overrightarrow{b}\times \overrightarrow{c}+y\overrightarrow{c}\times \overrightarrow{a}+z\overrightarrow{a}\times \overrightarrow{b}$ then

• A. $xy+yz+zx=0$
• B. $x=y=z$
• C. $x^3+y^3+z^3=3xyz$
• D. $x+y+z=1$

Answer 1

The correct option is C $x^3+y^3+z^3=3xyz$

$0=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{d}$
$=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).(x\overrightarrow{b}\times \overrightarrow{c}+y\overrightarrow{c}\times \overrightarrow{a}+z\overrightarrow{a}\times \overrightarrow{b})$
$=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})(x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c})$
$=(x+y+z)[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}];(x+y+z)[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=0$
But $[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]\ne 0$
$\therefore x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz$