Question

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

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  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
    #SPJ4

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