Answers:
(i) 3
(ii) [tex]-2x + 4y + 2z = 2[/tex]
Explanation:
(i) The area of the parallelogram bounded by vectors u and v is given by:
[tex]\text{Area} = \left \| u \times v \right \|[/tex]
Note that if [tex]u = (u_1, u_2, u_3) = (1, 0, 2), v = (v_1, v_2, v_3) = (0, -1, 2) [/tex],
[tex]u \times v
\\ = \left(u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1\right)
\\ = \left((0)(2)-(2)(-1), (2)(0)-(1)(2), (1)(-1)-(0)(0)\right)
\\ \boxed{u \times v = \left(2, -2, -1 \right)}[/tex]
Thus, the area of the parallelogram formed by vectors u and v is calculated as
[tex]\text{Area} = \left \| u \times v \right \|
\\ = \left \| \left(2, -2, -1 \right) \right \|
\\ = \sqrt{(2)^2 + (-2)^2 + (1)^2}
\\ \boxed{\text{Area} = 3}[/tex]
Hence, the area of the parallelogram is 3.
(ii) To obtain the equation of the plane, we must get its normal vector and a point in the plane.
As calculated in (i), the normal vector to the plane is [tex]v \times w[/tex] because the plane is parallel to vectors v and w. Note that if [tex]v = (v_1, v_2, v_3) = (0, -1, 2), w = (w_1, w_2, w_3) = (2, 1, 0)[/tex],
[tex]v \times w
\\ = \left(v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1\right)
\\ = \left((-2)(0)-(2)(1), (2)(2)-(0)(0), (0)(1)-(-1)(2)\right)
\\ \boxed{v \times w = \left(-2, 4, 2 \right)}[/tex]
Thus, a normal vector to the plane is [tex]\left(-2, 4, 2 \right)[/tex].
To get a point in the plane, note that vector u can be represented by any pairs of points. So, we can take (0, 0, 0) to be the starting point of vector u. Since [tex]u = (1, 0, 2)[/tex], the tip of vector u is [tex](0 + 1, 0, 0 + 2) = (1, 0, 2)[/tex]. Because the plane passes thru the tip of vector u, a point in the plane has coordinates [tex] (1, 0, 2)[/tex].
So for any point [tex](x, y, z)[/tex] in the plane, the equation is given by
[tex]n \cdot (x - 1, y - 0, z - 2) = 0[/tex]
Where n is the normal vector to the plane.
Since [tex]n = \left(-2, 4, 2 \right)[/tex], the equation becomes
[tex]\left(-2, 4, 2 \right) \cdot (x - 1, y - 0, z - 2) = 0
\\ \left(-2, 4, 2 \right) \cdot (x - 1, y, z - 2) = 0
\\ -2(x - 1) + 4y + 2(z - 2) = 0
\\ -2x + 2 + 4y + 2z - 4 = 0
\\ -2x + 4y + 2z - 2 = 0
\\ \boxed{-2x + 4y + 2z = 2 }[/tex]
