Chapter 1: Vectors, Lines and Planes
Section 1.5: Applications of Vector Products
Example 1.5.9
Derive the formula given in Table 1.5.1 for the distance from a point to a plane.
Solution
In Figure 1.5.9(a), the vectors A and B lie in the plane determined by the points Q, R, and S. (The positions of A and B have been reversed with respect to their locations in Figure 1.5.8(a).)
The blue vector is A×B, and is therefore normal to the plane. The dotted red line from the tip of C (which corresponds to point P) is parallel to the plane and delineates the projection of C onto A×B.
The distance from point P to the plane is the length of the projection of C onto A×B , a length designated by d, the leg of a right triangle whose hypotenuse is C.
Therefore, d=C cosθ, where θ is the angle between C and A×B.
Figure 1.5.9(a) Vectors A, B, C, A×B, and the projection of C onto A×B
Since cosθ = A×B·CA×B C,
d=CA×B·CA×B C = A×B·CA×B = A·B×C∥A×B∥ = ABC∥A×B∥
Note the use of the identity A·B×C=A×B·C that was established in Part (b) of Example 1.5.1. Also, there are orientations of A, B, and C for which ABC is negative, so the complete formula for d contains the absolute value of the box product in the numerator.
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