Show that a vector equation of a plane is r=(3i+2j)+s(i+3j+k)+t(2i-2j) has the line v=(3i+6j+k)+d(5i+7j+3k)?

Question

A hint on how to get started is appreciated as well

Show that a vector equation of a plane is r=(3i+2j)+ s (i+3j+k)+ t (2i-2j) has the line v=(3i+6j+k)+d(5i+7j+3k)?

Answer 1

Show that the vector equation of a plane
r = (3i + 2j) + s(i + 3j + k) + t(2i - 2j)

contains the line
v = (3i + 6j + k) + d(5i + 7j + 3k).

All we need to do is show that one point on the line also lies in the plane and that the directional vector of the line is perpendicular to the normal vector of the plane.

First let's find the normal vector n of the plane. The two directional vectors u and w that are in the plane are:

u = <1,3,1>
w = <2,-2,0>

u X w = 2i + 2j - 8k
Any multiple of that vector will do as well. Divide by 2.
n = i + j - 4k

If the line lies in the plane its directional vector will be perpendicular to n. So the dot product of the two vectors should be zero.

The directional vector of the line is <5,7,3>.

n•<5,7,3> = <1,1,-4>•<5,7,3> = 5 + 7 - 12 = 0

So the directional vector of the line lies in the plane. If we can also find a point in the line that lies in the plane we are done.

Set d = 0. One point on the line is (3,6,1).

First write out the equation of the plane in general form. We already have the normal vector to the plane.

n = i + j - 4k
r = (3i + 2j) + s(i + 3j + k) + t(2i - 2j)

Set s = t = 0 to find a point in the plane.

(x,y,z) = (3,2,0)

The equation of the plane is:

1(x - 3) + 1(y - 2) - 4(z - 0) = 0
x - 3 + y - 2 - 4z + 0 = 0
x + y - 4z - 5 = 0

A point on the line determined above is (3,6,1). Let's see if it is in the plane.

x + y - 4z - 5 = 0
3 + 6 - 4*1 - 5 = 0

So the line is in the plane.

Answer 2

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Answer 3

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