Question

Calculus III Questions:
1. Let P be the point (1, 2, 3, 4) and Q the point (4, 3, 2, 1). Let A be the vector <1, 1, 1, 1>. Let L be the line passing through P and parallel to A.
(a) Given a point X on the line L compute the distance between Q and X (as a function of parameter t).
(b) Show that there is precisely one point X0 on the line such that this distance achieves a minimum, and that this minimum is 2sqrt(5).
(c) Show that X0-Q is perpendicular to the line.

Answer 1

Help?

Answer 2

It's a textbook question in the chapter on differentiation of vectors. So I know I have to get the derivative somewhere, I just don't know how to approach the problem.

Answer 3

This is a bit complicated. First, the equation of the line,using vectors will be

L = P + tA (Caps will mean a vector, lower case a scalar)

(a) distance between points Q and X (X is on the line) is
| Q-X| = sqrt |Q-X|^2 = sqrt ( (Q-X)' (Q-X) ) where ' means transpose.
As the expression is unwieldy, let's find the length squared.
Because X is on the line, X= P + tA sub into the expression for magnitude squared

(Q-P -tA)' (Q-P -tA)

expand:
distance squared = Q'Q-P'Q-tQ'A-P'Q+P'P+tP'A-tA'Q+tA'P+t^2A'A
distance squared = Q'Q -2P'Q +P'P -2tQ'A +2t P'A + t^2 A'A
(using the fact that P'Q= Q'P and P'A= A'P)

(b) take the derivative with respect to t and set to zero, solve for t
X0= P+ t0 A where t0 is the solution just found.
find the length of vector Q-X0 Use |Q-X0|^2 =(Q-X0)' (Q-X0) = 20, and sqrt(20)= 2sqrt(5)

(c) Let Y = X0-Q, and show that Y'A = 0 (where A is the slope of the line)

Answer 4

I left a lot of the work unspecified. If you get stuck, ask more questions.

Answer 5

when you write ' means transpose, do you mean ' = derivative?

Answer 6

i've never seen the term "transpose"

Answer 7

no, I mean transpose of a vector. Normally you use T as an exponent

Answer 8

Do you know "dot product" of two vectors?

Answer 9

of course

Answer 10

when you multiply vectors , they have to have the same shape. In this case, P'Q (for example) means P dot Q

Answer 11

How would you multiply P times Q?

Answer 12

P*Q = <4, 6, 6, 4>
* = dot product

Answer 13

OK, that will work. Treat all the A'B as A dot B. Hopefully it makes sense

Answer 14

(a) makes sense
for (b) why does (Q-X0)*(Q-X0) = 20? where is 20 from

Answer 15

waaiit. nvm

Answer 16

(b) makes sense as well

Answer 17

(b) is where you have to do the work. That is the answer

Answer 18

and for c it's just use of the dot product because dot product = 0 means perpendicular.

Answer 19

how do you get A - the slope of the line for (c)

Answer 20

yes.
It isn't too hard to finish off (b) and (c) (I hope!)

Answer 21

rather, what equation am I getting the derivative of for (c)?

Answer 22

Let A be the vector <1, 1, 1, 1>. Let L be the line passing through P and parallel to A.

Answer 23

oh okay

Answer 24

cool!

Answer 25

no derivative for (a) or (c).

Answer 26

do you have time to answer another (less long) calc III question? it's hard to get good answerers for calc 3 stuff

Answer 27

Is it posted? If not, post it, because this one is long.

Answer 28

okay i'll make it a new post