(i) The foot of the perpendicular from the origin to a Plane is P(4, �

1.	(i) The foot of the perpendicular from the origin to a Plane is P(4, – 2,5). Write \((\bar{OP})\)(ii) Find the equation of the Plane in vector and Cartesian form.
Answer» (i) \(\overline{OP}\)= 4i - 2j + 5k (ii) Then is perpendicular unit vector to the required plane is \(\frac{\overline{OP}}{OP}=\frac{4i-2j+5k}{\sqrt{16+4+25}}\\=\frac{4i-2j+5k}{\sqrt{45}}\) The perpendicular distance from the origin is \(\sqrt{16+4+25}=\sqrt{45}\) Vector equation of the Plane can be written as \(\bar{r}.\bar{m}=d⇒\bar{r}.\frac{4i-2j+5k}{\sqrt{45}}\\=\sqrt{45}{}\) ⇒ \(\bar{r}.4i-2j+5k=45\) Cartesian from is 4x – 2y + 5z = 45

(i) The foot of the perpendicular from the origin to a Plane is P(4, – 2,5). Write \((\bar{OP})\)(ii) Find the equation of the Plane in vector and Cartesian form.

Answer»

(i) \(\overline{OP}\)= 4i - 2j + 5k

(ii) Then is perpendicular unit vector to the required plane is

\(\frac{\overline{OP}}{OP}=\frac{4i-2j+5k}{\sqrt{16+4+25}}\\=\frac{4i-2j+5k}{\sqrt{45}}\)

The perpendicular distance from the origin is

\(\sqrt{16+4+25}=\sqrt{45}\)

Vector equation of the Plane can be written as

\(\bar{r}.\bar{m}=d⇒\bar{r}.\frac{4i-2j+5k}{\sqrt{45}}\\=\sqrt{45}{}\)

⇒ \(\bar{r}.4i-2j+5k=45\)

Cartesian from is 4x – 2y + 5z = 45