Let the system of linear equations -x+2y-9z=7 -x+3y+7z=9 -2x+y+5z=8 -3x+y+13z= has a uniqu…

Question

Let the system of linear equations

$-x+2y-9z=7$

$-x+3y+7z=9$

$-2x+y+5z=8$

$-3x+y+13z=\lambda $

has a unique solution $x=\alpha ,y=\beta ,z=\gamma $. Then the distance of the point

$(\alpha ,\beta ,\gamma )$ from the plane $2x-2y+z=\lambda $ is :

Accepted Answer

$\begin{gathered} -x+2y-9z=7-(1) \ -x+3y-7z=9-(2) \ -2x+y+5z=8-(3) \ (2)-(1) \ y+16z=2\, (4) \ (3)-2	imes (1) \ -3y+23z=-6-(5) \ 3	imes (4)+(5) \ 71z=0\Rightarrow z=0 \ \, y=2 \ (-3,2,0)	o (\alpha ,\beta ,\gamma ) \ -3x+y+13z=\lambda \ \lambda =9+2=11 \end{gathered}$

The equation of the plane is:

$2x-2y+z=\lambda $

We already know that $\lambda =11$. So the equation of the plane becomes:

$2x-2y+z=11$

Now, let's find the distance $d$ of the point $(\alpha ,\beta ,\gamma )=(-3,2,0)$ from the plane using the formula :

$d=\frac{|Ax+By+Cz+D|}{\sqrt{{A}^{2}+{B}^{2}+{C}^{2}}}$

Plugging in the values for the point and the plane equation, we get:

$d=\frac{|2(-3)-2(2)+0-11|}{\sqrt{{2}^{2}+(-2{)}^{2}+{1}^{2}}}=\frac{|-6-4-11|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=7$

So, the distance of the point $(-3,2,0)$ from the plane $2x-2y+z=11$ is 7.

Solutions · 1