Question

Solve the following equations:
$x^2+y^2+z^2=21a^2,yz+zx-xy=6a^2,3x+y-2z=3a$.

Answer 1

$x^2+y^2+z^2=21a^2$ .......... $\left(i\right)$

$yz+zx-xy=6a^2$

$\Rightarrow 2(yz+zx-xy)=12a^2$ ......... $\left(ii\right)$

$3x+y-2z=3a$ ...... $\left(iii\right)$

$x^2+y^2+z^2=21a^2$ ........ $\left(iv\right)$
Subtracting $\left(iv\right)-\left(ii\right)$, gives
$x^2+y^2+z^2+2xy-2yz-2zx={9a}^2$

$\Rightarrow {(z-x-y)}^2={9a}^2$

$\Rightarrow z-x-y=\pm 3a$ ...... $\left(v\right)$

$\Rightarrow 2z-2x-2y=6a$ ......... $\left(vi\right)$ or $2z-2x-2y=-6a$...... $\left(vii\right)$

Adding $\left(iii\right)$ and $\left(vi\right)$, we get

$x-y=9a$ ........ $\left(vi\right)$
Adding $\left(iii\right)$ and $\left(vii\right)$, we get

$x-y=-3a$
Adding $\left(iii\right)$ and $\left(v\right)$, we get
$2x-z=6a$

$\Rightarrow z-2y=12a$ ............ $\left(viii\right)$ From $\left(vi\right)$

$\Rightarrow x^2+{(x-9a)}^2+{(2x-6a)}^2=21a^2$ ...... From $\left(i\right)$

$\Rightarrow x^2+x^2+81a^2-18ax+4x^2+36a^2-24ax=21a^2$

$\Rightarrow 6x^2-42ax+96a^2=0$

$\Rightarrow x^2-7ax+16a^2=0$ ..... $\left(ix\right)$

Adding $\left(iii\right)$ and $\left(v\right)$, we get
$2x-z=0$

$z=2x$

$x^2+{(x+3a)}^2+{\left(2x\right)}^2=21a^2$ ...... From $\left(i\right)$

$6x^2+6ax-12a^2=0$

$x^2+ax-2a^2=0$

$(x+2a)(x-a)=0$

$x=a,y=4a,x=2a$

$x=-2a,y=a,z=-4a$