Form line and find the parametric vector equation the line

MATH 241 Sections Exam 1 Spring 2021 (JWG)

Exam Submission:

5. Or you are welcome to write the answers on a separate piece of paper if other options don’t appeal to you, then scan and upload.

Exam Rules:

Work Shown:

1. Show all work as appropriate for and using techniques learned in this course.

Find the equation of the plane containing both the point and the line and write

it in the form ax + by + cz = d.

PQ = 1ˆı + 1 ˆ + 3ˆk

N =¯L × PQ = 3ˆı + 0 ˆ − 1ˆk

and (8, 10) form a line and find the parametric vector equation of the line.

Solution:

¯r(t) = tˆı +

�151 − 12t

�

		[10 pts]
		[10 pts]

Show that the line and the plane are parallel and find the distance between them.

Solution:

=\| − 17 + 7\|√6 = 10√6

(−1, 2, 1)

(1, 2, 11)

	¯r(t) = t2ˆı + (2t + 3) ˆ − tˆk x − 2y + z = 18	[5 pts]
		[5 pts]

t2− 2(2t + 3) − t = 18
t2− 5t − 24 = 0
(t − 8)(t + 3) = 0

Thus they meet at ¯r(−3) = 9ˆı − 3 ˆ + 3ˆk or (9, −3, 3) and they meet at¯r(8) = 64ˆı + 19 ˆ − 8ˆk or (64, 19, −8).

We have ¯r′(t) = 2tˆı + 2 ˆ − 1ˆk and so the distance is:		[5 pts]
� 8−3

6. Consider the curve with parameterization given here:

¯r(t) = (t4+ t)ˆı + 2t3ˆ − (t + 8)ˆk

	[5 pts]

(c)¯T (1) [5 pts] Solution:
We have:

¯v(1) ¯T (1) =			1 √62
=		6 √62ˆ −	1 √62

(d) a ¯T(1) [5 pts]

12√3 √62� 5 62ˆı +√
6

62ˆ −√
1

√62¯N(1) =�√12ˆı + 12 ˆ + 0ˆk

12√3(12ˆı + 12 ˆ + 0 ˆk ) − 132 3� 5 62ˆı +√62ˆ −√62

ˆk

�

7. Find the point on the sphere (x − 1)2+ (y + 2)2+ (z − 3)2= 16 which is as far as possible from the point (3, 5, 5). [10 pts]

	�	2ˆı + 7 ˆ + 2ˆk√57	�	=	8 √57ˆı + 28 57ˆ +	8 √57	ˆk

�1 −		8 √57