Mathematics: 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

halfStep-by-step explanation:...

Let u = (1,0,2) , v = (0,-1,2) and w = (2,1,0). - id1020122101460190, jeffreyaxtell4039

Ответ:

savannabroyhill24

15.03.2022

Mathematics

Answers:

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

Explanation:

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

$\text{Area} = \left \| u \times v \right \|$

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

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

,

$u \times v 
\\ = \left(u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1\right)
\\ = \left((0)(2)-(2)(-1), (2)(0)-(1)(2), (1)(-1)-(0)(0)\right)
\\ \boxed{u \times v = \left(2, -2, -1 \right)}$

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

$\text{Area} = \left \| u \times v \right \|
\\ = \left \| \left(2, -2, -1 \right) \right \|
\\ = \sqrt{(2)^2 + (-2)^2 + (1)^2}
\\ \boxed{\text{Area} = 3}$

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 $v \times w$ 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 = (v_1, v_2, v_3) = (0, -1, 2), w = (w_1, w_2, w_3) = (2, 1, 0)

,

$v \times w
\\ = \left(v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1\right) 
\\ = \left((-2)(0)-(2)(1), (2)(2)-(0)(0), (0)(1)-(-1)(2)\right)
\\ \boxed{v \times w = \left(-2, 4, 2 \right)}$

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

$\left(-2, 4, 2 \right) \cdot (x - 1, y - 0, z - 2) = 0 
\\ \left(-2, 4, 2 \right) \cdot (x - 1, y, z - 2) = 0 
\\ -2(x - 1) + 4y + 2(z - 2) = 0 
\\ -2x + 2 + 4y + 2z - 4 = 0 
\\ -2x + 4y + 2z - 2 = 0 
\\ \boxed{-2x + 4y + 2z = 2 }$

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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

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