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AlekseyPX
3 years ago
6

Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,

b = i + 2 j + 3 k, and c = 2 i + j + 2 k .
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

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

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