 # If p,q,r,x,y,z are non zero real number such that px+qy+rz+sqrt((p^2+q^2+r^2)(x^2+y^2+z^2))=0 Then (py)/(qx)+(qz)/(ry)+(rx)/(pz) is what try (px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2) p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2y^2+p^2z^2+q^2x^2+q^2y^2+q^2z^2+r^2x^2+r^2y^2+r^2z^2 2pqxy+2qryz+2prxz=p^2y^2+p^2z^2+q^2x^2+q^2z^2+r^2x^2+r^2y^2 How do i solve it Help me please Krish Crosby 2022-09-14 Answered
If p,q,r,x,y,z are non zero real number such that
$px+qy+rz+\sqrt{\left({p}^{2}+{q}^{2}+{r}^{2}\right)\left({x}^{2}+{y}^{2}+{z}^{2}\right)}=0$
Then $\frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is
what try
$\left(px+qy+rz{\right)}^{2}=\left({p}^{2}+{q}^{2}+{r}^{2}\right)\left({x}^{2}+{y}^{2}+{z}^{2}\right)$
${p}^{2}{x}^{2}+{q}^{2}{y}^{2}+{r}^{2}{z}^{2}+2pqxy+2qryz+2prxz={p}^{2}{x}^{2}+{p}^{2}{y}^{2}+{p}^{2}{z}^{2}+{q}^{2}{x}^{2}+{q}^{2}{y}^{2}+{q}^{2}{z}^{2}+{r}^{2}{x}^{2}+{r}^{2}{y}^{2}+{r}^{2}{z}^{2}$
$2pqxy+2qryz+2prxz={p}^{2}{y}^{2}+{p}^{2}{z}^{2}+{q}^{2}{x}^{2}+{q}^{2}{z}^{2}+{r}^{2}{x}^{2}+{r}^{2}{y}^{2}$
How do i solve it?
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Write $v=\left(p,q,r\right)$ and $w=\left(x,y,z\right)$. Then the given relation states
$v\cdot w+|v||w|=0$
But $v\cdot w=|v||w|\mathrm{cos}\theta$ where $\theta$ is the angle between them, so
$|v||w|\left(\mathrm{cos}\theta +1\right)=0$
Since none of the scalars are zero, we get $\mathrm{cos}\theta =-1$, so v=kw for some nonzero $k\in \mathbb{R}$ and
$\frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}=\frac{kxy}{kyx}+\frac{kyz}{kzy}+\frac{kzx}{kxz}=3$
###### Not exactly what you’re looking for? Kallie Fritz
By C-S
$0=px+qy+rz+\sqrt{\left({p}^{2}+{q}^{2}+{r}^{2}\right)\left({x}^{2}+{y}^{2}+{z}^{2}\right)}\ge px+qy+rz+|px+qy+rz|$
and since
$px+qy+rz+|px+qy+rz|\ge 0,$
we obtain
$px+qy+rz+|px+qy+rz|=0,$
which gives
$px+qy+rz\le 0.$
Also, the equality occurs for
$\left(x,y,z\right)||\left(p,q,r\right),$
which says that there is k<0, for which
$\left(p,q,r\right)=k\left(x,y,z\right).$
Thus,
$\sum _{cyc}\frac{py}{qx}=\sum _{cyc}\frac{kxy}{kyx}=3.$