Planes And Lines

[Aladdin]

Dear Staff , As usual I prefer hints and explanations on the way to approach the answer.

Consider the plane (P) of equation : x - 2y + 3z + 1 = 0.

1- Find two normal vectors to (P).
My Answer ; N1 ( 1 , -2 , 3 ) N2 ( -1 , 2 , -3)

2- Determine three points A , B and C on (P) .

3- Is the point I( 1 , -2 , 3 ) on (P) .

Thanks in advance ,
Aladdin

 

[soroban]

Hello, Aladdin!

Consider the plane \(\displaystyle P\) withf equation: .\(\displaystyle x - 2y + 3z + 1 \:=\: 0\)

1. Find two normal vectors to \(\displaystyle P.\)

\(\displaystyle \text{My answer }\quad N_1:\;( 1 , -2 , 3 ) \qquad N_2:\ -1 , 2 , -3)\)

Right!

\(\displaystyle \text{2. Determine three points }A , B\text{ and }C\text{ on }P.\)

\(\displaystyle \text{Let: }\:x=0,\:y=0\)

\(\displaystyle \text{The equation becomes: }\:0 - 0 + 3z +1\:=\:0 \quad\Rightarrow\quad z \:=\:-\tfrac{1}{3}\)

\(\displaystyle \text{Hence, we have: }\:A\left(0,\:0,\:-\tfrac{1}{3}\right)\qquad\hdots\;\;\text{Got it?}\)

\(\displaystyle \text{3. Is the point }I( 1 , -2 , 3 )\text{ on }P?\)

\(\displaystyle \text{Does }x=1,\:y=-2,\:z=3\text{ satisfy the equation?}\)

. . \(\displaystyle \text{Does: }\:1 - (-2) + 3(3) + 1 \:=\:0\;?\quad\hdots\:\text{ no}\)

\(\displaystyle \text{Therefore, point }I\text{ is }not\text{ on the plane.}\)

 

[Aladdin]

Thanks Soroban , I'm pleased with the way you answered my question .

Yes - I Got It ...