672 CHAPTER 9 Polar Coordinates; Vectors Proof of Property (8) Let = + + a b c u i j k 1 1 1 and = + + a b c v i j k. 2 2 2 Then ( ) ( ) ( ) × = − − − + − b c b c a c a c a b a b u v i j k 1 2 2 1 1 2 2 1 1 2 2 1 Now compute the dot product ( ) ⋅ × u u v . ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) ⋅ × = + + ⋅ − − − + − = − − − + − = a b c b c b c a c a c a b a b a b c b c b a c a c c a b a b u u v i j k i j k 0 1 1 1 1 2 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 1 2 2 1 1 1 2 2 1 Since two vectors are orthogonal if their dot product is zero, it follows that u and ×u v are orthogonal. Similarly, ( ) ⋅ × = v u v 0, so v and ×u v are orthogonal. ■ Figure 86(a) u v Figure 86(b) u v Finding a Vector Orthogonal to Two Given Vectors Find a vector that is orthogonal to = − + u i j k 32 and = − + − v i j k 3 . Solution EXAMPLE 4 Based on property (8), such a vector is ×u v. ( ) ( ) [ ] ( ) × = − − − =− −−−− +− =−++ u v i j k i j k i j k 3 2 1 1 3 1 2 3 3 1 9 2 2 7 The vector − + + i j k 2 7 is orthogonal to both u and v. Check: Two vectors are orthogonal if their dot product is zero. ( ) ( ) ( ) ⋅−+ + = − + ⋅−+ + =−−+= u i j k i j k i j k 2 7 3 2 2 7 3 4 7 0 ( ) ( ) ( ) ⋅−+ + =−+ − ⋅−+ + =+−= v i j k i j k i j k 2 7 3 2 7 1 6 7 0 Figure 87 v u u Finding the Area of a Parallelogram Find the area of the parallelogram whose vertices are ( ) = P 0, 0, 0 , 1 ( ) = − P 3, 2, 1 , 2 ( ) = − − P 1, 3, 1 , 3 and ( ) = P 2, 1, 0 . 4 EXAMPLE 5 4 Find a Vector Orthogonal to Two Given Vectors As long as the vectors u and v are not parallel, they will form a plane in space. See Figure 86(a). Based on property (8), the vector ×u v is normal to this plane. As Figure 86(a) illustrates, there are essentially (without regard to magnitude) two vectors normal to the plane containing u and v. It can be shown that the vector ×u v is the one determined by the thumb of the right hand when the other fingers of the right hand are cupped so that they point in a direction from u to v. See Figure 86(b).* Now Work PROBLEM 41 The proof of property (9) is left as an exercise. See Problem 62. 5 Find the Area of a Parallelogram Proof of Property (10) Suppose that u and v are adjacent sides of a parallelogram. See Figure 87. Then the lengths of these sides are u and v . If θ is the angle between u and v, then the height of the parallelogram is θ v sin and its area is θ [ ] = × = = × u v u v Area of parallelogram Base Height sin ↑ Property (9) ■ *This is a consequence of using a “right-handed” coordinate system.
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