Posted by **Shaila** on Monday, September 6, 2010 at 8:16pm.

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.

- Math - Equations of plane(check) -
**Reiny**, Monday, September 6, 2010 at 10:55pm
I actually did this problem

a) yes

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

- Math - Equations of plane(check) -
**Shaila**, Wednesday, September 15, 2010 at 7:13am
for the e part of the question you mean:

[x,y,z]=[2,3,1]+t[10,2,0]+s[5,1,0]

right??

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