35 POINTS!!!!!!!!!!!!!!!!!!!
1 answer:
You might be interested in
Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0
Answer:
48
Step-by-step explanation:
is basically the horizontal axis.
First, find the integral of x^2-25.
Remember that
integral of a constant is that constant times x.
Also that
to take the integral of a power function, add 1 to the degree and divide by that same degree.
We then get
Evaluate at -3
Then we evaluate at 0
Next, we subtract the the answer then we get
Given:
M is the midpoint of AB.
M(2,0) and A(-3, 3).
To find:
The coordinates of point B.
Solution:
Midpoint formula:
Let the coordinates of point B are (a,b). Then, using the midpoint formula, we get
On comparing both sides, we get
And,
Therefore, the coordinates of point B are (7,-3).