10. For the line L : r= (1,-5) + s(3,5), sER determine the following:
a. an equation for the line perpendicular to L, passing through P(2,0)
b. the point at which the line in part a. intersects the y-axis
So the first answer i got and it was r= (2,0) + t(2,3), tER
I don't know how to do second question, thanks for your help :)
(I tried doing it.. But i got the wrong answer, the book says the
answer is: (0,-1.2)
a. an equation for the line perpendicular to L, passing through P(2,0)
b. the point at which the line in part a. intersects the y-axis
So the first answer i got and it was r= (2,0) + t(2,3), tER
I don't know how to do second question, thanks for your help :)
(I tried doing it.. But i got the wrong answer, the book says the
answer is: (0,-1.2)
-
10. a.
The vector which multiplies the parameter - in this case (3,5) - is what determines the slope. To get a perpendicular line, swap the x and y and make one of them negative (this has the effect of new_slope = -1/old_slope); so it will be "something + s(-5,3)". Since it goes through (2,0), solution is:
r = (2,0) + s(-5,3)
or
r = (2,0) + s(5,-3)
(or t instead of s, doesn't matter).
b.
You want to find where x=0. Using
(2,0) + s(-5,3)
x=0 with s * -5 = -2, i.e. s = 2/5
(2,0) + 2/5(-5,3) = (0, 2/5 *3) = (0, 6/5) = (0, 1.2)
I'm very wary of going against your book but I've had a good look at this and I'm pretty sure I'm right.
The vector which multiplies the parameter - in this case (3,5) - is what determines the slope. To get a perpendicular line, swap the x and y and make one of them negative (this has the effect of new_slope = -1/old_slope); so it will be "something + s(-5,3)". Since it goes through (2,0), solution is:
r = (2,0) + s(-5,3)
or
r = (2,0) + s(5,-3)
(or t instead of s, doesn't matter).
b.
You want to find where x=0. Using
(2,0) + s(-5,3)
x=0 with s * -5 = -2, i.e. s = 2/5
(2,0) + 2/5(-5,3) = (0, 2/5 *3) = (0, 6/5) = (0, 1.2)
I'm very wary of going against your book but I've had a good look at this and I'm pretty sure I'm right.