To write the vector equation of a plane you need two different direction vectors on the plane, and a point on the plane
a) you are given exactly what I described above
so
r = (2,-4,6) + s(6,-7,0) + t(4,1,3)
It is easy to write these in parametric form
x = 2 + 6s + 4t
y = -4 -7t
z = 6 = 3t , it should be obvious what I did.
b) each of the lines gives you a direction vector, and you are given a point, so ....... (same question)
c) You can find vector AB and vector AC or vector BC , and you have the choice of 3 different points,
d) "has x-intercept 12, y-intercept -4, and z-intercept 7" translates to ...
you are given the points (12,0,0), (0,-4,0) and (0.0.7). Now it becomes the same kind of problem as c)
Hi, my exam is tomorrow and I'm here with my girlfriend who's also in the same class trying to figure out how to do the exam review but we're stuck.. please be kind and help us out, we'd really appreciate it. Thank you in advance.
QUESTION:
Write a vector equation and parametric equations for a plane that satisfies each set of conditions.
a) contains the point P̥(2,-4,6); has direction vectors a = [6,-7,0] and b = [4,1,3]
b) contains the point P̥(-5,9,-3); is parallel to the lines [x,y,z] = [1,-2,7] + s[4,-1,-3] and [x,y,z] = [7,-2,15] + t[1,6,-8]
c) contains the points A(4,-2,5), B(3,-3,1), and C(5,2,8)
d) has x-intercept 12, y-intercept -4, and z-intercept 7
please and thank you!
1 answer
1 answer