beljuA

2021-05-14

Find a nonzero vector orthogonal to the plane through the points P, Q, and R. and area of the triangle PQR
Consider the points below
P(1,0,1) , Q(-2,1,4) , R(7,2,7)
a) Find a nonzero vector orthogonal to the plane through the points P,Q and R
b) Find the area of the triangle PQR

### Answer & Explanation

saiyansruleA

Solution to your question with vector:

Jeffrey Jordon

a) The given vertices are $P\left(1,0,1\right),Q\left(-2,1,4\right)$ and $R\left(7,2,7\right)$ is

The normal vector to the plane passing through the given three points is $n=PQ×PR$

$=|\begin{array}{ccc}i& j& k\\ -3& 1& 3\\ 6& 2& 6\end{array}|$

$=\left(6-6\right)i-\left(-18-18\right)j+\left(-8-6\right)k$

$=\left(0\right)i+36j-12k$

$=<0,36,-12>$

This normal vector is orthogonal to the plane throught the point $P,Q,R$

b) $|PQ×PR|=\sqrt{{0}^{2}+{36}^{2}+\left(-12{\right)}^{2}}=\sqrt{1296+144}=\sqrt{1440}$

$=12\sqrt{10}$

Area of the triangle with verties $P,Q$ and $R$ is

$A=\frac{1}{2}|PQ×PR|$

$=\frac{1}{2}\left(12\sqrt{10}\right)$

$=6\sqrt{10}$

Jeffrey Jordon

a) To find a nonzero vector orthogonal to the plane through the points $P$, $Q$, and $R$, we can first find two vectors in the plane and take their cross product.
One way to do this is to find the vectors $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$, which lie in the plane.
$\stackrel{\to }{PQ}=\left[\begin{array}{c}-2-1\\ 1-0\\ 4-1\end{array}\right]=\left[\begin{array}{c}-3\\ 1\\ 3\end{array}\right]$ and $\stackrel{\to }{PR}=\left[\begin{array}{c}7-1\\ 2-0\\ 7-1\end{array}\right]=\left[\begin{array}{c}6\\ 2\\ 6\end{array}\right]$
Now we can take their cross product:
$\stackrel{\to }{PQ}×\stackrel{\to }{PR}=\left[\begin{array}{c}-3\\ 1\\ 3\end{array}\right]×\left[\begin{array}{c}6\\ 2\\ 6\end{array}\right]=\left[\begin{array}{c}0\\ -21\\ -9\end{array}\right]$
This vector is orthogonal to the plane through $P$, $Q$, and $R$. We can check that it is nonzero by calculating its length:
$|\stackrel{\to }{PQ}×\stackrel{\to }{PR}|=\sqrt{{0}^{2}+\left(-21{\right)}^{2}+\left(-9{\right)}^{2}}=\sqrt{450}=15\sqrt{2}$
So a unit vector orthogonal to the plane is
$\frac{\stackrel{\to }{PQ}×\stackrel{\to }{PR}}{|\stackrel{\to }{PQ}×\stackrel{\to }{PR}|}=\frac{1}{15\sqrt{2}}\left[\begin{array}{c}0\\ -21\\ -9\end{array}\right]=\left[\begin{array}{c}0\\ -\frac{7}{\sqrt{2}}\\ -\frac{3}{\sqrt{2}}\end{array}\right]$
b) The area of the triangle $PQR$ is half the length of the cross product of $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$.
$|\stackrel{\to }{PQ}×\stackrel{\to }{PR}|=\sqrt{{0}^{2}+\left(-21{\right)}^{2}+\left(-9{\right)}^{2}}=\sqrt{450}=15\sqrt{2}$
So the area of triangle $PQR$ is
$\frac{1}{2}|\stackrel{\to }{PQ}×\stackrel{\to }{PR}|=\frac{1}{2}\left(15\sqrt{2}\right)=\overline{)6\sqrt{10}}$

Vasquez

a) We can find the equation of the plane through $P$, $Q$, and $R$ by first finding two vectors in the plane.
$\stackrel{\to }{PQ}=\left[\begin{array}{c}-2-1\\ 1-0\\ 4-1\end{array}\right]=\left[\begin{array}{c}-3\\ 1\\ 3\end{array}\right]$ and $\stackrel{\to }{PR}=\left[\begin{array}{c}7-1\\ 2-0\\ 7-1\end{array}\right]=\left[\begin{array}{c}6\\ 2\\ 6\end{array}\right]$
Now we can find the normal vector to the plane by taking the cross product of $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$:
$\stackrel{\to }{n}=\stackrel{\to }{PQ}×\stackrel{\to }{PR}=\left[\begin{array}{c}-3\\ 1\\ 3\end{array}\right]×\left[\begin{array}{c}6\\ 2\\ 6\end{array}\right]=\left[\begin{array}{c}0\\ -21\\ -9\end{array}\right]$
This vector is orthogonal to the plane through $P$, $Q$, and $R$. We can check that it is nonzero by calculating its length:
$|\stackrel{\to }{n}|=\sqrt{{0}^{2}+\left(-21{\right)}^{2}+\left(-9{\right)}^{2}}=\sqrt{450}=15\sqrt{2}$
So a unit vector orthogonal to the plane is
$\frac{\stackrel{\to }{n}}{|\stackrel{\to }{n}|}=\frac{1}{15\sqrt{2}}\left[\begin{array}{c}0\\ -21\\ -9\end{array}\right]=\left[\begin{array}{c}0\\ -\frac{7}{\sqrt{2}}\\ -\frac{3}{\sqrt{2}}\end{array}\right]$
b) The area of the triangle $PQR$ is half the length of the cross product of $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$. We can use the equation we found for the normal vector to the plane to calculate the area:
$A=\frac{1}{2}|\stackrel{\to }{PQ}×\stackrel{\to }{PR}|=\frac{1}{2}|\stackrel{\to }{n}|=\frac{1}{2}\left(15\sqrt{2}\right)=\overline{)6\sqrt{10}}$
This method involves finding the normal vector to the plane first, and then using it to calculate the area.

RizerMix

Result:
$6\sqrt{\left(}10\right)$
Explanation:
a) To find a vector orthogonal to the plane through the points $P,Q$ and $R$, we can take the cross product of two non-parallel vectors that lie in the plane. We can choose the vectors $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$, which are given by:
$\stackrel{\to }{PQ}=\left(\begin{array}{c}-2-1\\ 1-0\\ 4-1\end{array}\right)=\left(\begin{array}{c}-3\\ 1\\ 3\end{array}\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\stackrel{\to }{PR}=\left(\begin{array}{c}7-1\\ 2-0\\ 7-1\end{array}\right)=\left(\begin{array}{c}6\\ 2\\ 6\end{array}\right).$
Then, the vector orthogonal to the plane is given by their cross product:
$\stackrel{\to }{n}=\stackrel{\to }{PQ}×\stackrel{\to }{PR}=\left(\begin{array}{c}-3\\ 1\\ 3\end{array}\right)×\left(\begin{array}{c}6\\ 2\\ 6\end{array}\right)=\left(\begin{array}{c}-20\\ 0\\ 20\end{array}\right).$
Therefore, a nonzero vector orthogonal to the plane through $P,Q$ and $R$ is $\overline{)\stackrel{\to }{n}=\left(-20,0,20\right)}$.
b) The area of the triangle $PQR$ is half the magnitude of the cross product of $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$:
$\text{Area}\left(PQR\right)=\frac{1}{2}‖\stackrel{\to }{PQ}×\stackrel{\to }{PR}‖=
\frac{1}{2}‖\left(\begin{array}{c}-3\\ 1\\ 3\end{array}\right)×\left(\begin{array}{c}6\\ 2\\ 6\end{array}\right)‖=
\frac{1}{2}‖\left(\begin{array}{c}-20\\ 0\\ 20\end{array}\right)‖=
\frac{1}{2}·20\sqrt{{2}^{2}+{0}^{2}+{2}^{2}}=
\overline{)6\sqrt{10}}.$

Therefore, the area of the triangle $PQR$ is $\overline{)6\sqrt{10}}$.

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