Question

Let

Answer 1

Answer:

3.3

Step-by-step explanation:

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

$u_1\times u_2$

$\begin{pmatrix} - 2\\ - 4\\ 1 \end{pmatrix}\times\begin{pmatrix} - 4\\ 1 \\ - 4\end{pmatrix}=\begin{pmatrix} 15 \\ - 12\\ - 18\end{pmatrix}$

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

$d=\dfrac{|Ax_1+By_1+Cz_1 +D|}{\sqrt{A^2+B^2+C^2}}$

Comparing,

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

Substituting,

$d=\dfrac{|(15\times3)+(-12\times-8)+(-18\times3)+0|}{\sqrt{15^2+(-12)^2+(-18)^2}}$

$d=\dfrac{|45+96-54|}{\sqrt{225+144+324}}=\dfrac{87}{\sqrt{693}}$

$d =\dfrac{87}{26.32\ldots}\approx 3.3$