Question

chegg Find an equation of the plane. the plane through the point (5, 2, 4) and with normal vector 8i 3j 4k

1. An equation of the plane. the plane through the point (5, 2, 4) and with normal vector 8i + 3j + 4k are-
• Cartesian form: 8x + 3y + 4z = 72
• Vector form: r(8i + 3j + 4k) = 72

### What is normal vector?

A normal vector would be a vector that forms a 90° angle with that other surface, vector, or axis and is perpendicular to those objects.
• Each vector has a direction and a certain magnitude. The largest component of a vector in mathematics is its magnitude, though often magnitude is not all that relevant.
• The requirement determines everything. Sometimes all we need is some guidance. Because of this, magnitude is not required in these situations.
• As a result, we can assert that a vector’s direction is special. A normal vector to plane is located upon that line, and there are numerous vectors on that alignment which are normal to the surface.
• This idea may also be seen graphically. Therefore, direction adds individuality to the system.
calculation for the equation;
Normal vector =>  8i + 3j + 4k
Points through which plane is passing are (5,2,4)
As the plane which is normal is passing through the given plane, it must satisfy the plane;
Thus,
8(x-5) + 3(y-2) + 4(z-4) = 0
8x – 40 + 3y – 6 + 4z -16 = 0
8x + 3y + 4z – 72 = 0
8x + 3y + 4z = 72 ( equation of plane is Cartesian form)
The equation of plane in vector form is;
r(8i + 3j + 4k) = 72
To know more about the normal vector, here
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