Question

Write the equation of the plane with normal vector n passing through point P, in the scalar form ax + by + cz = d. n= (-6,2,5), Po = (4,9,5) (Express numbers in exact form. Use symbolic notation and fractions where needed.) Write the equation of the plane with normal vector n = 4j passing through the point (6, 12,9) in scalar form ax + by + cz = d. (Express numbers in exact form.

Answer 1

To find the equation of a plane, one uses the normal vector and a point from the plane in the scalar form ax + by + cz = d. By substituting the point into this form, we obtain the specific equation for each plane.

Step-by-step explanation:

To find the equation of the plane with a given normal vector n and passing through a specific point P, we use the scalar form ax + by + cz = d. The coefficients a, b, and c are the components of the normal vector, and d is calculated by plugging the point's coordinates into the equation.

For the first part, the normal vector n is (-6, 2, 5), and the point Po is (4, 9, 5). To find the equation, we use these in the scalar form:

-6x + 2y + 5z = d.

We determine d by substituting the coordinates of point Po into the equation:

-6(4) + 2(9) + 5(5) = d

d = -24 + 18 + 25

d = 19

So, the plane's equation is -6x + 2y + 5z = 19.

For the second part, the normal vector is given as 4j, which in component form is (0, 4, 0). The point is (6, 12, 9), so the equation becomes:

0x + 4y + 0z = d.

Plugging in the point's coordinates gives:

0(6) + 4(12) + 0(9) = d

d = 48

Thus, the plane's equation is y = 12 or, in scalar form, 0x + 4y + 0z = 48.