Chapter 1: Vectors, Lines and Planes
Section 1.5: Applications of Vector Products
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Example 1.5.9
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Derive the formula given in Table 1.5.1 for the distance from a point to a plane.
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Solution
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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).)
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The blue vector is , 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 .
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The distance from point P to the plane is the length of the projection of C onto , a length designated by , the leg of a right triangle whose hypotenuse is .
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Therefore, , where is the angle between C and .
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Figure 1.5.9(a) Vectors A, B, C, , and the projection of C onto
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Since = ,
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= = =
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Note the use of the identity that was established in Part (b) of Example 1.5.1. Also, there are orientations of A, B, and C for which is negative, so the complete formula for contains the absolute value of the box product in the numerator.
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