Question

In Exercises 36–41 use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

$P(3,1,8),Q(0,6,-1),R(-3,5,-3)$

Short answer

(a) A nonzero vector N perpendicular to the plane determined by the points are $(-19,-1,10)$.

(b) Two unit vectors perpendicular to the plane determined by the points are $\pm \frac{1}{\sqrt{462}}(-19,-1,10)$.

(c) The area of the triangle determined by the points is $\frac{\sqrt{462}}{2}$.

Step-by-step solution

Step 1. Given Information

In the given exercises use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

The given points are$P(3,1,8),Q(0,6,-1),R(-3,5,-3)$

Part (a) Step 1. firstly finding a nonzero vector N perpendicular to the plane determined by the points.

We have$P(3,1,8),Q(0,6,-1),R(-3,5,-3)$

Now

role="math" localid="1649672822896" $$\begin{gathered} \overrightarrow{PQ}=(0-3,6-1,-1-8)=(-3,5,-9) \\ \overrightarrow{PR}=(-3-3,5-1,-3-8)=(-6,4,-11) \end{gathered}$$

Part (a) Step 2. Now finding PQ→×PR→

$$\begin{gathered} \overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccc}i& j& k\\ -5& 5& -9\\ -6& 4& -11\end{array}\right| \\ \overrightarrow{PQ}\times \overrightarrow{PR}=i\left|\begin{array}{cc}5& -9\\ 4& -11\end{array}\right|-j\left|\begin{array}{cc}-5& -9\\ -6& -11\end{array}\right|+k\left|\begin{array}{cc}-5& 5\\ -6& 4\end{array}\right| \\ \overrightarrow{PQ}\times \overrightarrow{PR}=i\{5\times (-11)-4\times (-9\left)\right\}-j\left\{\right(-5)\times (-11)-(-9)\times (-6\left)\right\}+k\left\{\right(-5)\times 4-5\times (-6\left)\right\} \\ \overrightarrow{PQ}\times \overrightarrow{PR}=i(-55+36)-j(55-54)+k(-20+30) \\ \overrightarrow{PQ}\times \overrightarrow{PR}=-19i-j+10k \end{gathered}$$

The points are$(-19,-1,10)$

Part (b) Step 1. Now finding two unit vectors perpendicular to the plane determined by the points. 

So,

$$\begin{gathered} ||\overrightarrow{PQ}\times \overrightarrow{PR}||=\sqrt{{(-19)}^2+{(-1)}^2+{\left(10\right)}^2} \\ ||\overrightarrow{PQ}\times \overrightarrow{PR}||=\sqrt{361+1+100} \\ ||\overrightarrow{PQ}\times \overrightarrow{PR}||=\pm \sqrt{462} \end{gathered}$$

Required vector

$$\begin{gathered} \frac{\overrightarrow{PQ}\times \overrightarrow{PR}}{||\overrightarrow{PQ}\times \overrightarrow{PR}||}=\frac{(-19,-1,10)}{\pm \sqrt{462}} \\ \frac{\overrightarrow{PQ}\times \overrightarrow{PR}}{||\overrightarrow{PQ}\times \overrightarrow{PR}||}=\pm \frac{1}{\sqrt{462}}(-19,-1,10) \end{gathered}$$

Part (c) Step 1. Now finding the area of the triangle determined by the points. 

$$\begin{gathered} \mathrm{Area}∆ABC=\frac{1}{2}||\overrightarrow{PQ}\times \overrightarrow{PR}|| \\ \mathrm{Area}∆ABC=\frac{1}{2}||\sqrt{462}|| \\ \mathrm{Area}∆ABC=\frac{\sqrt{462}}{2} \end{gathered}$$