Math 211 * Final Exam * Dec 20, 2017 * Solution * Victor Matveev (a) Since the line belongs to both planes, it is normal to both normal vectors, therefore its direction vector is: 1 1 2 2 4, 1, 2 4 1 2 3, 6, 9 3 1, 2, 3 1, 2, 1 1 2 1 i j k n v n n n A scalar factor does not change the vector’s direction, so for the sake of simplicity use v = 1, 2, 3To find a point on this line, consider its intersection with any coordinate plane, for instance x = 0: 0 0 0 2 2 3 1 1 0 2 , , 2 y z y x y z z r Thus, the line of intersection is: 0 2 1 2 0, 1, 1, 2, 3 2 3 t t t t x t y z t r r v Check your answers!! 4 2 4 (1 2) 2( 2 3) 3 :) 2 2 1 2 2 3 0 :) x y z t t t x y z t t t (b) The distance between a point and a line is illustrated below: 0 0 ˆ sin d PP r r v where 0 2,1, 1 1, 0, 1 1,1, 2 r r and 2, 2, 1 2, 2, 1 ˆ 3 4 4 1 v 0 1 1 25 9 16 ˆ 1 1 3 2 5, 3, 4 3 3 50 5 2 3 3 2 2 1 d i j k r r v
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Math 211 * Final Exam * Dec 20, 2017 * Solution * Victor Matveev
(a) Since the line belongs to both planes, it is normal to both normal vectors, therefore its direction vector is:
1
1 2
2
4, 1, 24 1 2 3, 6, 9 3 1, 2, 3
1, 2, 11 2 1
i j kn
v n nn
A scalar factor does not change the vector’s direction, so for the sake of simplicity use v = 1, 2, 3
To find a point on this line, consider its intersection with any coordinate plane, for instance x = 0:
000 2
2 31
10
2, ,
2
y z yx
y z z
r
Thus, the line of intersection is: 0 2 1 20, 1, 1, 2, 3
2 3
t t t t
x t
y
z t
r r v
Check your answers!!
4 2 4 ( 1 2 ) 2( 2 3 ) 3 :)
2 2 1 2 2 3 0 :)
x y z t t t
x y z t t t
(b) The distance between a point and a line is illustrated below:
0 0 ˆsind P P r r v
where 0 2, 1, 1 1, 0, 1 1, 1, 2r r and
2, 2, 1 2, 2, 1ˆ
34 4 1v
0
1 1 25 9 16ˆ 1 13
2 5, 3, 43 3
50 5 2
332 2 1
d
i j k
r r v
Note: there is a typo in this problem, which however does not affect the results: the constant should be e – 3, not 2e – 3