Find the equation of the plane pi through A(1,-1, 0) and perpendicular to the line L_1 :(x + 1)/4= (y + 2)/-5= (z + 3)/6. Also, find the distance between pi and the line L_2 : (x - 1)/5= y/4, z = 3.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the equation of the plane passing through the point A(1,-1,0) and perpendicular to the line `(x + 1)/4= (y + 2)/-5= (z + 3)/6` , hence, you need to consider the point `B(x,y,z)` in the plane `pi` , such that the line AB needs to be perpendicular to the direction vector of the line `L_1` , such that:

`bar v*bar (AB) = 0`

`bar v` represents the direction vector of the line `L_1`

`bar v = <4,-5,6>`

`bar (AB) = <x - x_A, y - y_A, z - z_A>`

`bar (AB) = <x - 1, y + 1, z>`

`<4,-5,6>*<x - 1, y + 1, z> = 0 => 4(x - 1) - 5(y + 1) + 6z =0`

`4x - 4 - 5y - 5 + 6z = 0`

`4x - 5y + 6z - 9 = 0`

Hence, evaluating the equation of the plane pi, under the given conditions, yields `(pi): 4x - 5y + 6z - 9 = 0.`

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