Asked by Angel
Find the scalar equation for the plane passing through the point P=(0, 2, −1) and containing the line L defined by
x = 2+t
y = 4+t
z = −6+7t
x = 2+t
y = 4+t
z = −6+7t
Answers
Answered by
Arora
Since t is a variable, you can obtain three different points on the plane:
a) (0,2,-1) Given in the question
b) (2,4,-6) by setting t=0
c) (3,5,1) by setting t=1
If you know three points in a plane, you can use them to find a unique equation for the plane. Try it out
a) (0,2,-1) Given in the question
b) (2,4,-6) by setting t=0
c) (3,5,1) by setting t=1
If you know three points in a plane, you can use them to find a unique equation for the plane. Try it out
Answered by
JAGJIT
Thank you Arora, It works first find the two vectors with the above points:
(0,2,-1)-(2,4,-6) = (-2,-2,5)
[3,5,1]-[2,4,-6] = [1,1,7]
cross multiply the points (to get normal vector) and we get:
19i+19j+0k
plug in the numbers: 19x+19y+0z=d
find d: 19(0)+19(2)=d ; d = 38
final solution: 19x+19y=38
(0,2,-1)-(2,4,-6) = (-2,-2,5)
[3,5,1]-[2,4,-6] = [1,1,7]
cross multiply the points (to get normal vector) and we get:
19i+19j+0k
plug in the numbers: 19x+19y+0z=d
find d: 19(0)+19(2)=d ; d = 38
final solution: 19x+19y=38
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