Question

1. let $p(1,2,3), q(1, - 1, - 2)$, and $r(0,0,0)$ be three points in $r^{3}$. (a) find an equation of the plane through $p, q$, and $r$. (b) find the area of the triangle formed by $pqr$. (c) find the equation of the line though $p$ that is perpendicular to the plane from (a).

Explanation:

Step1: Find two vectors in the plane

The vectors $\overrightarrow{RP}=\langle1,2,3
angle$ and $\overrightarrow{RQ}=\langle1, - 1,-2
angle$.

Step2: Find the normal vector of the plane

The normal vector $\vec{n}=\overrightarrow{RP}\times\overrightarrow{RQ}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&2&3\\1&-1&-2\end{vmatrix}$$

=\vec{i}(-4 + 3)-\vec{j}(-2 - 3)+\vec{k}(-1 - 2)=-\vec{i}+5\vec{j}-3\vec{k}=\langle - 1,5,-3
angle$.
Since the plane passes through the origin $R(0,0,0)$, the equation of the plane is $-1(x - 0)+5(y - 0)-3(z - 0)=0$, or $-x + 5y-3z = 0$.

Step3: Find the area of the triangle

The area of the triangle with two - side vectors $\overrightarrow{RP}$ and $\overrightarrow{RQ}$ is $A=\frac{1}{2}\vert\overrightarrow{RP}\times\overrightarrow{RQ}\vert$.
$\vert\overrightarrow{RP}\times\overrightarrow{RQ}\vert=\sqrt{(-1)^2+5^2+(-3)^2}=\sqrt{1 + 25+9}=\sqrt{35}$. So $A=\frac{\sqrt{35}}{2}$.

Step4: Find the equation of the perpendicular line

The direction vector of the line perpendicular to the plane is the normal vector of the plane $\vec{n}=\langle - 1,5,-3
angle$, and the line passes through $P(1,2,3)$.
The parametric equations of the line are $x = 1-t,y = 2 + 5t,z = 3-3t$.

Answer:

(a) $-x + 5y-3z = 0$
(b) $\frac{\sqrt{35}}{2}$
(c) $x = 1-t,y = 2 + 5t,z = 3-3t$