Question

Let

yequals=left bracket Start 3 By 1 Matrix 1st Row 1st Column 3 2nd Row 1st Column negative 8 3rd Row 1st Column 3 EndMatrix right bracket



3


−8


3


​,



Bold u 1u1equals=left bracket Start 3 By 1 Matrix 1st Row 1st Column negative 2 2nd Row 1st Column negative 4 3rd Row 1st Column 1 EndMatrix right bracket



−2


−4


1


​,



Bold u 2u2equals=left bracket Start 3 By 1 Matrix 1st Row 1st Column negative 4 2nd Row 1st Column 1 3rd Row 1st Column negative 4 EndMatrix right bracket



−4


1


−4


.



Find the distance from y to the plane in



set of real numbers R cubedℝ3



spanned by



Bold u 1u1



and



Bold u 2u2.

Answer 1

3.3

Explanation:

We determine the plane formed by
`u_1` and
`u_2`. The normal to the plane is given by their cross product:

`u_1* u_2`

`\begin{pmatrix} - 2\\ - 4\\ 1 \end{pmatrix}*\begin{pmatrix} - 4\\ 1 \\ - 4\end{pmatrix}=\begin{pmatrix} 15 \\ - 12\\ - 18\end{pmatrix}`

The equation of the plane is then given by

`15x-12y-18z=0`

The distance between a vector
`\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}` and a plane
`Ax+By+Cz +D =0` is given by

`d=(|Ax_1+By_1+Cz_1 +D|)/(√(A^2+B^2+C^2))`

Comparing,

`A = 15,\ B = - 12,\ C = -18,\ D = 0,\ x_1=3,\ y_1 =-8, \ z_1 =3`

Substituting,

`d=(|(15*3)+(-12*-8)+(-18*3)+0|)/(√(15^2+(-12)^2+(-18)^2))`

`d=(|45+96-54|)/(√(225+144+324))=(87)/(√(693))`

`d =(87)/(26.32\ldots)\approx 3.3`