Question

Let u = (1,0,2) , v = (0,-1,2) and w = (2,1,0). Compute

(i) the area of the parallelogram bounded by u and v (5)

(ii) the equation of the plane parallel to v and w pass through the tip of u

Answer 1

Answers:

(i) 3
(ii)
`-2x + 4y + 2z = 2`

Explanation:

(i) The area of the parallelogram bounded by vectors u and v is given by:


`\text{Area} = \left \| u * v \right \|`

Note that if
`u = (u_1, u_2, u_3) = (1, 0, 2), v = (v_1, v_2, v_3) = (0, -1, 2)`,


`u * 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 * v = \left(2, -2, -1 \right)}`

Thus, the area of the parallelogram formed by vectors u and v is calculated as


`\text{Area} = \left \| u * v \right \| \\ = \left \| \left(2, -2, -1 \right) \right \| \\ = √((2)^2 + (-2)^2 + (1)^2) \\ \boxed{\text{Area} = 3}`

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
`v * w` because the plane is parallel to vectors v and w. Note that if
`v = (v_1, v_2, v_3) = (0, -1, 2), w = (w_1, w_2, w_3) = (2, 1, 0)`,


`v * 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 * w = \left(-2, 4, 2 \right)}`

Thus, a normal vector to the plane is
`\left(-2, 4, 2 \right)`.

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
`u = (1, 0, 2)`, the tip of vector u is
`(0 + 1, 0, 0 + 2) = (1, 0, 2)`. Because the plane passes thru the tip of vector u, a point in the plane has coordinates
`(1, 0, 2)`.

So for any point
`(x, y, z)` in the plane, the equation is given by


`n \cdot (x - 1, y - 0, z - 2) = 0`

Where n is the normal vector to the plane.

Since
`n = \left(-2, 4, 2 \right)`, the equation becomes


`\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 }`