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

Question

Let the system of linear equations

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

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

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

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

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

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

Accepted Answer

$$
\begin{aligned}
& -x+2 y-9 z=7-(1) \\
& -x+3 y-7 z=9-(2) \\
& -2 x+y+5 z=8-(3) \\
& (2)-(1) \\
& y+16 z=2 \quad(4) \\
& (3)-2 \times(1) \\
& -3 y+23 z=-6-(5) \\
& 3 \times(4)+(5) \\
& 71 z=0 \Rightarrow z=0 \\
& \quad y=2 \\
& (-3,2,0) \rightarrow(\alpha, \beta, \gamma) \\
& \text { Put in }-3 x+y+13 z=\lambda \\
& \lambda=9+2=11
\end{aligned}
$$

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