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find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).​

Sagot :

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).​

To find:

The component equation of the plane.

Solution:

The equation of plane is

[tex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[/tex]

Where, [tex](x_0,y_0,z_0)[/tex] is the point on the plane and [tex]\left< a,b,c\right>[/tex] is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

[tex]-2(x-(-10))+5(y-7)+1(z-5)=0[/tex]

[tex]-2(x+10)+5y-35+z-5=0[/tex]

[tex]-2x-20+5y-35+z-5=0[/tex]

[tex]-2x+5y+z-60=0[/tex]

[tex]-2x+5y+z=60[/tex]

Therefore, the equation of the plane is [tex]-2x+5y+z=60[/tex].

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