hw1.3.18

# hw1.3.18 - c This has no solution so L is Neither parallel...

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(week 3, continued) Problem 1.3.18: The line L passes through the point P = (1 , - 1 , 1) and has direction vector ± d = [2 , 3 , - 1] . For each of the following planes, determine whether L and the plane are parallel, perpendicular or neither. Solution: If the plane has normal vector ±n, the answer depends on whether ± d and ±n are parallel, perpendicular or neither. We explain below. (a) 2x+3y-z=1. Soln: ±n = [2 , 3 , - 1] and for ± d = [2 , 3 , - 1] ±n · ± d = [2 , 3 , - 1] · [2 , 3 , - 1] = 4 + 9 + 1 ± = 0 . This says that L is NOT Parallel to the plane. (why?) In fact, ±n = ± d, so L is Perpendicular to the plane. (b) 4 x - y + 5 z = 0 has ±n = [4 , - 1 , 5] and ±n · ± d = [4 , - 1 , 5] · [2 , 3 , - 1] = 8 - 3 - 5 = 0 , so L IS Parallel to this plane. (why?)

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2 (c) x - y - z = 3 has ±n = [1 , - 1 , - 1] and ±n · ± d = [1 , - 1 , 1] · [2 , 3 , - 1] = 2 - 3 - 1 ± = 0 , so L isn’t parallel. Now we check c±n = ± d = [1 , - 1 , 1] , so c [1 , - 1 , 1] = [2 , 3 , - 1] , so we need c = 2 , - c = 3 and c = - 1 , all for the same
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Unformatted text preview: c. This has no solution, so L is Neither parallel nor perpendicular to this plane. Last comments on Homework 2 (I hope!): From our text we have the statement Problem 1.2.34, Section 1.2, pg. 26: Find the projection of ±v onto ±u where ±u = [1 ,-1] , ±v = [3 ,-1] . We recall (from your homework papers) that proj ±u ( ±v ) was a vector in the direction of ±u = [1 ,-1] . As a clariﬁcation, we note Example, NOT Problem 1.2.34: Find the projection of ± b = [1 ,-1] onto ±a = [3 ,-1] , for which we found proj ±a ( ± b ) = 4 10 ±a = 2 5 [3 ,-1] . For emphasis, this vector ±a = [3 ,-1] was NOT the text’s vector ±u (and WORSE, was the text’s vector ±v . . . )....
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hw1.3.18 - c This has no solution so L is Neither parallel...

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