Find the linear equation of the plane through the point (6,−2,5) and with normal

tabita57i 2021-09-17 Answered
Find the linear equation of the plane through the point (6,−2,5) and with normal vector j+2k.

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Expert Answer

Leonard Stokes
Answered 2021-09-18 Author has 16468 answers
Step 1
Given, point (6,-2,5) and normal vector j+2k
Step 2
Concept:
When point \(\displaystyle{\left({x}_{{{0}}},{y}_{{{0}}},{z}_{{{0}}}\right)}\) and normal vector ai+bj+ck, then equation of plane is given by \(\displaystyle{a}{\left({x}-{x}_{{{0}}}\right)}+{b}{\left({y}-{y}_{{{0}}}\right)}+{c}{\left({z}-{z}_{{{0}}}\right)}={0}\)
Step 3
Here, \(\displaystyle{\left({x}_{{{0}}},{y}_{{{0}}},{z}_{{{0}}}\right)}={\left({6},-{2},{5}\right)}\) and a=0, b=1, c=2
Then, equation of plane is 0(x-6)+1(y-(-2))+2(z-5)=0
(y+2)+2z-10=0
Equation of plane y+2z-8=0
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Answered 2021-11-27 Author has 10829 answers

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