SOLUTION: Please explain how to solve this: https://docs.google.com/document/d/1SspsOLwJsYo44NsPnNKjAf7lmEUuKn9YK1oUL4PiE9k/edit?usp=sharing

Algebra ->  Vectors -> SOLUTION: Please explain how to solve this: https://docs.google.com/document/d/1SspsOLwJsYo44NsPnNKjAf7lmEUuKn9YK1oUL4PiE9k/edit?usp=sharing      Log On


   

Question 1189173: Please explain how to solve this:
https://docs.google.com/document/d/1SspsOLwJsYo44NsPnNKjAf7lmEUuKn9YK1oUL4PiE9k/edit?usp=sharing
Found 2 solutions by Edwin McCravy, rothauserc:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The line l1 has eq:

r%22%22=%22%22%28matrix%283%2C1%2C3%2C1%2C2%29%29%22%22%2B%22%22t%2A%28matrix%283%2C1%2C-1%2C3%2C2%29%29

(a)  the point A(-8,34,n) lies on l1, find the value of n.

r%22%22=%22%22%28matrix%283%2C1%2Cx%2Cy%2Cz%29%29%22%22=%22%22%28matrix%283%2C1%2C3%2C1%2C2%29%29%22%22%2B%22%22t%2A%28matrix%283%2C1%2C-1%2C3%2C2%29%29

We substitute A for < x,y,z >

%28matrix%283%2C1%2C-8%2C34%2Cn%29%29%22%22=%22%22%28matrix%283%2C1%2C3%2C1%2C2%29%29%22%22%2B%22%22t%2A%28matrix%283%2C1%2C-1%2C3%2C2%29%29    system%28-8=3-t%2C34=1%2B3t%2Cn=2%2B2t%29

Both the first and second equations give t=11, 
substitute t=11 in the 3rd equation and get n=24.

--------------------------------------

(b) The line r%22%22=%22%22%28matrix%283%2C1%2C1%2C-2%2Cu%29%29%22%22%2B%22%22s%2A%28matrix%283%2C1%2C-1%2Cp%2Cq%29%29

intersects l1 at A and is perpendicular to l1.

We know this second line also goes through A, so we substitute A

%28matrix%283%2C1%2C-8%2C34%2C24%29%29%22%22=%22%22%28matrix%283%2C1%2C1%2C-2%2Cu%29%29%22%22%2B%22%22s%2A%28matrix%283%2C1%2C-1%2Cp%2Cq%29%29

I'm tired.  Maybe I'll finish later.  You have to make sure the dot product is zero.


Edwin

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
8.a Since the point (-8, 34, n) lies on I(1), we know that
;
-8 = 3 -t
34 = 1+3t
n = 2 +2t
;
3 -t = -8
-t = -11
t = 11
:
n = 2 +2(11) = 24
;
8.b Two vectors are perpendicular if their dot product is 0
;
We are given that the lines intersect at point A, then
;
-8 = 1 -s
34 = -2 +s*p
24 = u +s*q
;
1 -s = -8
s = 9
:
34 = -2 +9p
9p = 36
p = 4
;
24 = u +9q
;
Let t = 0 and s = 0, then
;
vector v = <3-(-8), 1-34, 2-24> = <11, -33, -22>
:
vector w = <1-(-8), -2-34, u-24> = <9, -36, u-24>
;
The dot product of vectors w and v is 0, therefore
;
9 * 11 +(-33)*(-36) +(-22)*(u-24) = 0
;
99 +1188 -22u +528 = 0
;
-22u +1815 = 0
;
u = 82.5
:
We know that
;
24 = u +9q
:
82.5 +9q = 24
;
9q = -58.5
;
q = -6.5
:
Therefore,
;
p = 4, q = -6.5, u = 82.5
;