+0

0
99
1

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 $$d$$?

Nov 8, 2019

#1
+109064
+1

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

Nov 8, 2019
edited by CPhill  Nov 8, 2019