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

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

for the e part of the question you mean:

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

right??

Your answers are partially correct. Here are the correct responses:

a) To find two vectors parallel to the plane, you can take any two points on the plane and subtract their coordinates to get the two vectors. So, in this case, AB = B - A = (-11, 1, 2) - (2, 3, 1) = (-13, -2, 1), and AC = C - A = (-7, -3, -6) - (2, 3, 1) = (-9, -6, -7).

b) To find two vectors perpendicular to the plane, you need to find the normal vector of the plane. One way to do this is to take the cross product of the two vectors you found in part a. So, the normal vector N = AB x AC = (-13, -2, 1) x (-9, -6, -7) = (-44, 116, -64).

c) To write a vector equation of the plane, you can use one of the points on the plane (let's use point A) and the normal vector N. So, the vector equation of the plane is (x, y, z) = (2, 3, 1) + t(-44, 116, -64), where t is any real number.

d) To write the scalar equation of the plane, you can use the normal vector N and any point on the plane (such as point A). The equation of the plane is given by N · (x, y, z) = N · A, where · represents the dot product. So, (-44, 116, -64) · (x, y, z) = (-44, 116, -64) · (2, 3, 1).

e) To find an equation of the line through the x- and y-intercepts of the plane, you need to find the x- and y-intercepts. For the x-intercept, set z = 0 and solve for x and y. For the y-intercept, set z = 0 and solve for x and y. These two points will define the line, and you can express it in parametric form using a parameter t.

a) Find two vectors that are parallel to the plane:

The vectors AC, BC, or AB will indeed be parallel to the plane. Since A, B, and C are points on the plane, any vector connecting two of these points will lie entirely in the plane, making it parallel.

b) Find two vectors that are perpendicular to the plane:
To find vectors perpendicular to the plane, you can calculate the cross product of any two non-parallel vectors within the plane. For example, you can take the cross product of vectors AB and AC to get a vector perpendicular to the plane.

c) Write a vector equation of the plane:
To write a vector equation of the plane, you can choose any point on the plane, let's say A(2,3,1) and determine two vectors within the plane (e.g., AB and AC). The vector equation of the plane can be written as:
P = A + s*(AB) + t*(AC), where s and t are scalar parameters.

d) Write the scalar equation of the plane:
To write the scalar equation of the plane, you can determine the normal vector of the plane by calculating the cross product of vectors AB and AC. Let's say the normal vector is N. The scalar equation of the plane can be written as:
N · (P - A) = 0, where · denotes the dot product.

e) Write an equation of the line through the x- and y-intercepts of the plane:
To find the x- and y-intercepts of the plane, you can set z = 0 in the vector equation of the plane. Let's say the x-intercept is (x0, 0, 0) and the y-intercept is (0, y0, 0). Then the equation of the line passing through these intercepts can be written as:
x = x0 + t*(x0 - 0)
y = y0 + t*(y0 - 0)
z = 0

Your answers are correct! Good job. Let me know if you have any further questions.