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

Find the linear equation of the plane through the point (6,−2,5) and with normal vector j+2k.

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Leonard Stokes
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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