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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
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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

 

 

cool cool cool

 Nov 8, 2019
edited by CPhill  Nov 8, 2019

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