# write in equation of plane with points A(2,-1,3),B(3,1,2)/plane paralell to vector v=(3,-1,-4)

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### 1 Answer

The vector orthogonal to the plane that passes through A and B is called normal vector: `bar n ` = (a,b,c).

Since the vector `bar v ` is parallel to the plane comprising the points A and B, then is `bar n ` is perpendicular to `bar v` .

The cross product of bar n and bar v yields 0 and the cross products of `bar n` and `bar (AB)` yields 0.

`bar n*bar v = a*3 + b*(-1) + c*(-4) =gt bar n*bar v = 3a - b - 4c = 0`

`bar n*bar (AB) = a*(3-2) + b*(1+1) + c*(2-3) =gt bar n*bar (AB) = a + 2b - c = 0 =gt a = c - 2b`

Plugging a in the equation 3a - b - 4c = 0 yields:

`3c - 6b - b - 4c = 0 =gt -c - 7b = 0 =gt c = -7b =gt b = -c/7`

`a = c + c/7 = 8c/7`

Taking c = 7 yields:

a = 8 and b = -1

The normal vector n is bar n = (8 ; -1 ; 7)

The equation of the plane is 8(x - 2) - (y + 1) + 7(z - 3) = 0 => 8x - 16 - y - 1 + 7z - 21 = 0.

Adding like terms yields: 8x - y + 7z - 38 = 0

**Hence, evaluating the equation of the plane yields: 8x - y + 7z - 38 = 0.**