Consider the plane that contains points A(2,3,1), B(-11,1,2), C(-7,-3,-6).
a) Find two vectors that are parallel to the plane.
Ans: AC, BC or AB will be parallel to the plane.
b) Find two vectors that are perpendicular to the plane.
Ans: if i find the normal vector then i can use dot product and use arbitrary values to find the two vectors
c) Write a vector equation of the plane
Ans: I have two points AC, BC or AB and i have the normal, so i use them to write the vector equation.
d) Write the scalar equation of the plane.
Ans: the normal vectors are my coefficients for the(x,y,z) therefore if i substitute a point to find the constant 'd'
e) Write an equation of the line through the x- and y- intercepts of the plane
Ans: Find the x and y intercepts and those will be my points my z value being zero and using the normal as direction vector
Are the answers correct? I am a little confused, please correct if i am wrong somewhere.
I actually did this problem
b) the simplest normal is [1,-5,3) so any other multiple of that would do
c) [x,y,z] = [2,3,1] + s[13,2,-1] + t[9,6,7]
d) I got x - 5y + 3z = -10 , all 3 original points satisfied
e) for x-intercept, let both y and z = 0, so x=-10
for the point (-10,0,0)
for y-intercept let both x and z = 0, y = 2
for the point (0,2,0)
direction vector of line joining the intercepts is [10, 2,0] or [5,1,0]
so equation of line is [x,y,z] = 0,2,0) + t[5,1,0]
check my arithmetic, last time I made a silly error if you recall
posted by Reiny
for the e part of the question you mean:
right??posted by Shaila