Recall that an equation for the yz plane is x 0 if x

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Recall that an equation for the yz-plane is x = 0. If (x 0 ,y 0 ,z 0 ) is the point of intersection, then there is a number t 0 so that 12 + 4t 0 = x 0 = 0 since (x 0 ,y 0 ,z 0 ) lies on both the plane and the line. Solving this linear thing yields t 0 = -3. Consequently, (x 0 ,y 0 ,z 0 ) = (0,-21,0).

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TEST1/MAC2313 Page 2 of 5 _________________________________________________________________ 5. (5 pts.) What point (x 0 ,y 0 ) is seven-tenths of the way from P = (-2,-3) to Q = (48,7) ?? The vector v with initial point P and terminal point Q has coordinates <50,10> in standard position. Consequently, the terminal point of <x 0 ,y 0 > = <-2,-3> + (7/10)<50,10> = <33,4> in standard position provides us with the coordinates for (x 0 ,y 0 ). Vector magic!! _________________________________________________________________ 6. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,2,4>. Then v w = (-3)(-1) + (-2)(2) + (1)(4) = 3 - 4 + 4 = 3 _________________________________________________________________ 7. (5 pts.) Suppose v = <-3,-2, 1> and w = <-2,2,1>. Then v × w = <-3,-2, 1> × <-2,2,1> = < -4, -(-1), -10> _________________________________________________________________ 8. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,2,4>. Then proj w ( v ) = <-1/7,2/7,4/7> , and the component of v perpendicular to w is w 2 = <-20/7,-16/7,3/7> . _________________________________________________________________ Here is the back of Page 4 of 5:
TEST1/MAC2313 Page 3 of 5 _________________________________________________________________ 9. (5 pts.) Suppose v = <-3,-2, 1> and w = <-2,2,1>. If α , β , and γ are the direction angles of w , then cos( α ) = -2/(9) 1/2 = -2/3 , cos( β ) = 2/(9) 1/2 = 2/3 ,and cos( γ ) = 1/(9) 1/2 = 1/3 . _________________________________________________________________ 10. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,2,4>. What is the exact value of the angle θ between v and w ?? θ = cos -1 (( v w )/( v w )) = cos -1 ( 3/( (14) 1/2 (21) 1/2 ) = cos -1 ( 3/(294) 1/2 ) = cos -1 ( 3/[7(6) 1/2 ] ) _________________________________________________________________ 11. (5 pts.) Write a point-normal equation for the plane perpendicular to v = <-3,-2,1> and containing the point (3 π ,2 π , π ). -3(x - 3 π ) - 2(y -2 π ) + (z - π ) = 0 _________________________________________________________________ 12. (5 pts.) Write an equation for the plane passing through the point (4,-5, 6) and perpendicular to the line defined by the vector equation <x,y,z> = <4e,25,2 π > + t<2,-3,-3>. The direction vector in the equation for the line provides us with a normal vector for the plane because the line is to be perpendicular to the plane.

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