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What is the relationship of the given line L : (x, y, z) = (1,-2,-3) + t<-4,-1,5>...

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gomezzzzzzzzz... | Student | eNoter

Posted February 4, 2013 at 5:46 AM via web

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What is the relationship of the given line L : (x, y, z) = (1,-2,-3) + t<-4,-1,5> to each of the planes below?

Plane contain L, plane is parallel to L or plane interests L?

2. -2x+5y+2z = 4

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 7, 2013 at 6:25 PM (Answer #1)

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You need to test if the line is parallel to the plane, hence, you need to evaluate the dot product of the direction vector of the line and the normal vector to the plane, such that:

`<-4,-1,5>*<-2,5,2> = (-4)*(-2) + (-1)*(5) + 5*2`

`<-4,-1,5>*<-2,5,2> = 8 - 5 + 10`

`<-4,-1,5>*<-2,5,2> = 13 != 0`

Since the dot product is not zero, hence, the direction vector is not perpendicular to the normal vector, thus, the line is not parallel to the plane.

You need to test if the direction vector is parallel to normal vector, hence, you need to check if the direction vector and normal vector are scalar multiples of each other, such that:

`<-4,-1,5> = k*<-2,5,2> => {(-2k = -4),(5k = 1),(2k = 5):}`

`{(k = 2),(k = 1/5),(k = 5/2):}`

Since the direction vector is not parallel to normal vector, hence, the line is not perpendicular to the plane.

Since the line intersects the plane, you need to find the coordinates of the intersection point, hence, you need to substitute x,y,z coordinates of direction vector of the line in equation of the plane, such that:

`-2( - 4t + 1)+5(-2 - t)+2(-3 + 5t) = 4`

`8t - 2 - 10 - 5t - 6 + 10t = 4 `

`13t = 18 + 4 => t = 22/13`

You need to substitute `22/13` for t in x,y,z coordinates of direction vector, such that:

`x = - 4(22/13) + 1 => x = (-88 + 13)/13 => x = -75/13`

`y = -2 - 22/13 => y = -48/13`

`z = -3 + 110/13 => z = (-39 + 110)/13 => z = 71/13`

Hence, the given line intersect the plane at the point `(-75/13,-48/13,71/13).`

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