## pottersheep 2 years ago Grade 12 Vectors . Find a point P on the line L situated 19√2 units distance from plane W.? Find a point P on the line L situated 19√2 units distance from plane W. W = 5x + 3y - 4z = 18 L = (x-1)/2 = (y+2)/-1 = z/-3 Please help me. How do I do this? Thank you so much

1. amistre64

find the point on the plane that the xyz of the line fits

2. pottersheep

For paramatric equation of the line I got X = 1+2t Y = -2 –t Z = -3t

3. amistre64

the normal of the plane is <5,3,-4> a line with that normal is X = 1+5t Y = -2 +3t Z = 0 -4t would you agree?

4. pottersheep

yes

5. amistre64

|dw:1371056632838:dw|

6. amistre64

im thinking we need to go another route, find the angle between the normal and the line

7. pottersheep

We don't deal with angles really..we use this formula though http://easycalculation.com/analytical/pointplane.gif

8. amistre64

i can never remember the formula, so i tend to go a longer route.

9. amistre64

if we plug in xyz from the line eq, into the plane we get: 5(1+2t)+3(-2-t)-4(-3t) = 18 5+10t -6-3t +12t = 18 19t = 19, when t=1 ---------------------- X = 1+2 = 3 Y = -2–1 = -3 Z = -3 so the point (3,-3,-3) is the point on the plane that the line peirces thru and the angle between the normal and line is: arccos(19/10sqrt(7)

10. amistre64

|dw:1371057451397:dw|

11. amistre64

if we attach the point in the plane to the line vector .. we get an equivalent setup for the line to play with x = 3 +2t y = -3 -1t z = -3 -3t we can use t as a scalar, using the appropriate "length" of the traingle

12. amistre64

19*sqrt(2) = 19sqrt(2) 10sqrt(7)*sqrt(2) = 10sqrt(14) = t to define the point is what im thinking

13. pottersheep

Thanks to what you did, I figured out what I think is another way, and got t = 11. can I type it out for you?

14. amistre64

by all means :)

15. pottersheep

For paramatric equation of the line I got X = 1+2t Y = -2 –t Z = -3t So any point on the line is (1+2t, -t-t, -3t) If I plug it into the equation for finding the distance of a point to a plane then 19/√2 = 5x +3y-4z – 18/(√50) 19/√2 = 5(1+2t) + 3 (-2 –t) -4(-3) = 18/(√50) 19/√2= (-19 + 19t)/√50 19 √100 = (-19 + 19t) 190 = (-19 + 19t) T = 11 

16. pottersheep

Once again thank you so much for your help, you're many people's hero :)

17. amistre64

youre welcome :)