Question

The plane passes through D(3,5,−4) with normal n=(2,−1,3). Determine the Scalar Form of the plane and the exact intercepts.

Answer 1

The scalar form of the plane passing through D(3,5,−4) with normal n=(2,−1,3) is 2x - y + 3z + 11 = 0. The exact intercepts are x = -5.5, y = 11, and z = -11/3.

Step-by-step explanation:

The scalar form of a plane equation is given by the formula: Ax + By + Cz + D = 0, where (A, B, C) is the normal vector to the plane, and D is a scalar. For a plane passing through point D(3,5,−4) with normal n=(2,−1,3), we can determine the scalar D as follows:

• Plug the coordinates of D into the plane equation: 2(3) - 1(5) + 3(-4) + D = 0.
• Solve for D: 6 - 5 - 12 + D = -11, thus D = 11.

Therefore, the scalar form of the plane is 2x - y + 3z + 11 = 0.

To find the intercepts, set two variables to zero and solve for the third one:

• x-intercept: Set y = 0, z = 0, resulting in 2x = -11, or x = -5.5.
• y-intercept: Set x = 0, z = 0, resulting in -y = -11, or y = 11.
• z-intercept: Set x = 0, y = 0, resulting in 3z = -11, or z = -11/3.