S6.194.4.12.4.29 Find a nonzero vector orthogonal to plane

[karush]

$\tiny{s6.194.4.12.4.29}$
$\textsf{a. Find a nonzero vector orthogonal to plane
through the points: }$
$\textsf{b. Find the area of the triangle PQR}$
\begin{align} \displaystyle
&P(1,0,0)& &Q(0,2,0)& &R(0,0,3)\\
%&=\color{red}{\frac{1209}{28} }
\end{align}
$\textit{do what first?}$

 

[Joppy]

$\tiny{s6.194.4.12.4.29}$
$\textsf{a. Find a nonzero vector orthogonal to plane
through the points: }$
$\textsf{b. Find the area of the triangle PQR}$
\begin{align} \displaystyle
&P(1,0,0)& &Q(0,2,0)& &R(0,0,3)\\
%&=\color{red}{\frac{1209}{28} }
\end{align}
$\textit{do what first?}$

You could start by drawing the vectors on a graph maybe (this will help with the second part too). This will give you a more intuitive understanding of what's happening. Next, you need to think about what the question means by how it says 'orthogonal'. Do you know what this means and how it applies to the question?

 

[karush]

orthogonal means perpendicular

 

[Joppy]

In finding a non-zero vector orthogonal to the plane we are essentially defining the plane. So, let's find the equation of the plane given the three points P, Q and R.

Let $\hat{n} = \hat{PR} \times \hat{PQ}$ be the vector normal to the plane.

$\hat{PQ} = \hat{OQ} - \hat{OP} = (0\hat{\imath} +2 \hat{\jmath} + 0\hat{k}) - (\hat{\imath} + 0\hat{\jmath} + 0\hat{k}) = -\hat{\imath} +2\hat{\jmath}$

$\hat{PR} = \hat{OR} - \hat{OP} = (0\hat{\imath} +0\hat{\jmath} + 3\hat{k}) - (\hat{\imath} + 0\hat{\jmath} + 0\hat{k}) = -\hat{\imath} + 3\hat{k}$

Now,

$\hat{n} = \hat{PR} \times \hat{PQ} =

\begin{vmatrix}\
\hat{\imath}&\hat{\jmath}&\hat{k} \\
-1&2&0 \\
-1&0&3
\end{vmatrix} = \hat{\imath} \begin{vmatrix} 2&0\\ 0&3 \end{vmatrix} - \hat{\jmath} \begin{vmatrix} -1&0\\ -1&3 \end{vmatrix} + \hat{k} \begin{vmatrix} -1&0\\ -2&3 \end{vmatrix}$

Can you finish it off?

 

[Greg]

In finding a non-zero vector orthogonal to the plane we are essentially defining the plane. So, let's find the equation of the plane given the three points P, Q and R.

Let $\hat{n} = \hat{PR} \times \hat{PQ}$ be the vector normal to the plane.

$\hat{PQ} = \hat{OQ} - \hat{OP} = (0\hat{\imath} +2 \hat{\jmath} + 0\hat{k}) - (\hat{\imath} + 0\hat{\jmath} + 0\hat{k}) = -\hat{\imath} +2\hat{\jmath}$

$\hat{PR} = \hat{OR} - \hat{OP} = (0\hat{\imath} +0\hat{\jmath} + 3\hat{k}) - (\hat{\imath} + 0\hat{\jmath} + 0\hat{k}) = -\hat{\imath} + 3\hat{k}$

Now,

$\hat{n} = \hat{PR} \times \hat{PQ} =

\begin{vmatrix}\
\hat{\imath}&\hat{\jmath}&\hat{k} \\
-1&2&0 \\
-1&0&3
\end{vmatrix} = \hat{\imath} \begin{vmatrix} 2&0\\ 0&3 \end{vmatrix} - \hat{\jmath} \begin{vmatrix} -1&0\\ -1&3 \end{vmatrix} + \hat{k} \begin{vmatrix} -1&0\\ -2&3 \end{vmatrix}$

Can you finish it off?

That should be

$$\begin{vmatrix}\
\hat{\imath}&\hat{\jmath}&\hat{k} \\
-1&2&0 \\
-1&0&3
\end{vmatrix} = \hat{\imath} \begin{vmatrix} 2&0\\ 0&3 \end{vmatrix} - \hat{\jmath} \begin{vmatrix} -1&0\\ -1&3 \end{vmatrix} + \hat{k} \begin{vmatrix} -1&2\\ -1&0 \end{vmatrix}$$

 

[karush]

(6,3,2)

 

[Greg]

Yes. The required area is 7/2. Do you see why?

 

[Joppy]

Whoops sorry about that. I also forgot to post this website for the OP. A nice simple way for plotting vectors. It also shows addition, subtraction, and cross product of 2 vectors.

 

[karush]

Whoops sorry about that. I also forgot to post this website for the OP. A nice simple way for plotting vectors. It also shows addition, subtraction, and cross product of 2 vectors.

yeah very cool I'll start using it here...better than W|A

 

[Joppy]

yeah very cool I'll start using it here...better than W|A

- - - Updated - - -Maybe not better, but definitely easier ;) :).

 

[karush]

Yes. The required area is 7/2. Do you see why?

TBH no

 

[Greg]

$\triangle{PQR}=\sqrt{6^2+3^2+2^2}/2=\sqrt{49}/2=7/2$