Let \(Q = (-2, 3, 4).\) Then the distance between \(Q\) and the plane through the points \(A = (0,1,1), B = (1,1,0)\) and \(C = (1,0,3)\) is \(\dfrac{d}{\sqrt{11}}\) for some value of \(d\). What's \(d\)?
+128631 We need to first find the eqution for the plane
Let vector AB = (1, 0 , -1)
Let vector AC = ( 1, - 1, 2 )
We can find the normal vector to the plane by taking the cross-product of AB x AC
n = i j k i j
1 0 -1 1 0 =
1 -1 2 1 -1
[ i * 0 * - 1 + j * -1 * 1 + k * 1 * - 1 ] - [ k * 0 * 1 + i * -1 * - 1 + j * 2 * 1 ] =
[ 0 - 1j - 1k ] - [ 0 + 1i + 2j ] = - 1i -3j - 1k
The equation of the plane using point A is
-1( x - 0) -3(y -1) - 1( z - 1) = 0
-1x - 3y + 3 - 1z + 1 = 0
-1x - 3y - 1z + 4 = 0
1x + 3y +1z - 4 = 0
The distance between Q and the plane is given by
l 1(-2) + 3(3) + 1( 4) - 4 l l 7 l 7
_______________________ = ______ = ___
√ [ 1^2 + 3^2 + 1^2 ] √11 √11
So d = 7
