Question

Find the equation of the plane through the origin and parallel to the plane 10x+3y+7z=6

Blank 1. Calculate the answer by read surrounding text.

x+ Blank 2. Calculate the answer by read surrounding text.

y+ Blank 3. Calculate the answer by read surrounding text.

z= Blank 4. Calculate the answer by read surrounding text.

Answer 1

Solution

The given plane is

$10x+3y+7z=6$

Its normal vector is

$\vec{n_{1}}=[10, 3, 7]$

We know that if the two planes are parallel, then they have the same normal vector or their normal vectors are parallel (i.e they are integral multiple of each other)

Therefore, the normal vector of the required plane is

$\vec{n_{2}}=[10, 3, 7]$

Now equation of the plane, passing through origin (0, 0, 0) with normal vector

$\vec{n_{2}}=[10, 3, 7]$ is

$(x-0)(10)+(y-0)(3)+(z-0)(7)=0$

$10x+3y+7z=0$

A plot of the required plane is shown below

Blank 1 --- 10

Blank 2 --- 3

Blank 3 --- 7

Blank 4 --- 0