Find an equation of the plane E containing the lines and..Linear algebra?

Question

Find an equation of the plane E containing the lines and
1(t)=(2–1t)i+(2+2t)j+(1–1t)k
I2(t)=(0+2t)i+(–1+3t)j+(–2+3t)k
E: ___x+___y+___z=___

Answer 1

Find an equation of the plane E containing the lines:

L1(s) = (2 - s)i + (2 + 2s)j + (1 - s)k
L2(t) = (0 + 2t)i + (-1 + 3t)j + (-2 + 3t)

I have changed the t in the first equation to s, because the t's in the two equations were not the same.
____________

First check to see if the lines are parallel or if they intersect. If neither of those is true, then the two lines cannot be contained in one plane.

The directional vectors are not non-zero multiples of each other, so the lines are not parallel. Check to see if they intersect by solving for s and t.

x = 2 - s = 2t
y = 2 + 2s = -1 + 3t
z = 1 - s = -2 + 3t

It turns out the planes intersect when
s = 0
t = 1

at the point P(2, 2, 1).

So the problem has a solution.

The normal vector n, of the plane containing the given lines will be normal to the directional vectors of both lines. Take the cross product.

n = <-1, 2, -1> X <2, 3, 3> = <9, 1, -7>

With the normal vector n and the point P in the plane we can write the equation of the plane.

9(x - 2) + 1(y - 2) - 7(z - 1) = 0
9x - 18 + y - 2 - 7z + 7 = 0
9x + y - 7z - 13 = 0
9x + y - 7z = 13

Answer 2

9x + y - 7z = 13

There are several ways to do this. Here's a fairly quick method.

Find three points not on the same line:
L1(0) = {2, 2, 1}
L1(1) = {1, 4, 0}
L2(0) = {0, -1, -2}

So if ax+by+cz=d, then
(1) 2a + 2b + c = d
(2) a + 4b = d
(3) -b - 2c = d

Subtract the twice the second from the first
-6b + c = -d
-12b + 2c = -2d
Add the third
-13b = -d
b = d/13
Then a = 9d/13 from (2) above
and c = -7d/13 from (3) above

So (9d/13)x + (d/13)y - (7d/13)z = d
Divide both sides by d.
(9/13)x + (1/13)y - (7/13)z = 1
or
9x + y - 7z = 13

Answer 3

get the cross product of the direction vectors of the two lines... that will be the normal vector of the plane.

choose any point on any line.

you can get E from the point-normal vector form of the plane.

Method:
.. <-1, 2, -1>
x < 2, 3, 3>

= <6+3, 3-2 , -3-4> = <9, 1, -7>

thus 9x + y - 7z = d

at the point(0, -1, -2) ... -1 + 14 = d = 13

thus 9x + y - 7z = 13


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Answer 4

First we make positive that strains might want to be positioned on one airplane. this happens if the strains are parallel (which they are not) or in the experience that they intersect. to locate factor of intersection, locate fee of t the position t = 3t, 2t = t, 3t = 8t ----> t = 0 -----> factor of intersection = (0, 0, 0) strains have route vectors < a million, 2, 3 > and < 3, a million, 8 > Vector it extremely is prevalent to airplane will be perpendicular to route vectors of line. to locate vector it extremely is perpendicular to those 2 vectors, in basic terms take pass product: < a million, 2, 3 > x < 3, a million, 8 > = < 13, a million, -5 > Now we may be able to locate equation of airplane given a level (0, 0, 0) and prevalent < 13, a million, -5 > 13 (x - 0) + a million (y - 0) - 5 (z - 0) = 0 13x + y - 5z = 0

Answer 5

complex step. do a search on a search engine. that may help!