hvacwk

2021-12-30

(a) Find the area of the triangle having vertices A(1, 0, 1), B(0, 2, 3), and C (2, 1, 0).
(b) Use the result of part (a) to find the length of the altitude from vertex C to side AB.

Elois Puryear

Step 1
Given:
A(1,0,1) , B(0,2,3) ,C(2,1,0)
Step 2
$\stackrel{\to }{AB}=\stackrel{\to }{B}-\stackrel{\to }{A}=\left(0,2,3\right)-\left(1,0,1\right)=\left(1,-2,-2\right)$
$\stackrel{\to }{BC}=\stackrel{\to }{C}-\stackrel{\to }{B}=\left(2,1,0\right)-\left(0,2,3\right)=\left(-2,1,3\right)$
$\stackrel{\to }{AB}×\stackrel{\to }{BC}=\left[\begin{array}{ccc}i& j& k\\ 1& -2& -2\\ -2& 1& 3\end{array}\right]$
$=i\left(-6+2\right)-j\left(3-4\right)+k\left(1-4\right)$
$=-4i+j-3k$
Area of triangle $=\frac{1}{2}|\stackrel{\to }{AB}×\stackrel{\to }{BC}|$
$=\frac{1}{2}\sqrt{{\left(-4\right)}^{2}+{\left(1\right)}^{2}+{\left(-3\right)}^{2}}$
$=\frac{1}{2}\sqrt{16+1+9}$
$=\frac{\sqrt{26}}{2}$
Step 3
Distance between $AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}$
$=\sqrt{{\left(0-1\right)}^{2}+{\left(2-0\right)}^{2}+{\left(3-1\right)}^{2}}$
$=\sqrt{1+4+4}$
$=3$
Area of triangle $=\frac{\sqrt{26}}{2}$
$\frac{1}{2}×base×height=\frac{\sqrt{26}}{2}$
$⇒\frac{1}{2}×3×height=\frac{\sqrt{26}}{2}$
$⇒height=\frac{\sqrt{26}}{3}$

Neil Dismukes

(a) We have $\stackrel{\to }{AB}=B-A=\left(0,2,3\right)-\left(1,0,1\right)=\left(-1,2,2\right)$
and $\stackrel{\to }{AC}=C-A=\left(2,1,0\right)-\left(1,0,1\right)=\left(1,1,-1\right)$
Then we have $\stackrel{\to }{AB}×\stackrel{\to }{AC}=\left(-1,2,2\right)×\left(1,1,-1\right)$
$=\left(\left[\begin{array}{cc}2& 2\\ 1& -1\end{array}\right],-\left[\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right],\left[\begin{array}{cc}-1& 2\\ 1& 1\end{array}\right]\right)$
$=\left(-4,1,-3\right)$
Hence the area of the triangle with the vertices A, B and C is given by:
$\frac{1}{2}||\stackrel{\to }{AB}×\stackrel{\to }{AC}||=\frac{1}{2}||\begin{array}{ccc}-4& 1& -3\end{array}||=\frac{1}{2}\sqrt{{\left(-4\right)}^{2}+{1}^{2}+{\left(-3\right)}^{2}}=\frac{\sqrt{26}}{2}$
(one half of the area of the parallelogram determined by ).
(b) Let h be the length of the altitude from C to AB. Then the area of the triangle with the vertices A, B and C is $\frac{1}{2}||\stackrel{\to }{AB}||h$. Therefore, by using part (a) we get $\frac{1}{2}||\stackrel{\to }{AB}||h=\frac{\sqrt{26}}{2}$
Also, we have $||\stackrel{\to }{AB}||=||\begin{array}{ccc}-1& 2& 2\end{array}||=\sqrt{{\left(-1\right)}^{2}+{2}^{2}+{2}^{2}}=\sqrt{9}=3$
This implies $h=\frac{\sqrt{26}}{||\stackrel{\to }{AB}||}=\frac{\sqrt{26}}{3}$

user_27qwe

$\begin{array}{}\text{(a) Following example 10,}\\ A=\frac{1}{2}||\stackrel{\to }{AB}×\stackrel{\to }{AC}||\\ \stackrel{\to }{AB}=\left(0,2,3\right)-\left(1,0,1\right)\\ =\left(-1,2,2\right)\\ \stackrel{\to }{AC}=\left(2,1,0\right)-\left(1,0,1\right)\\ =\left(1,1,-1\right)\\ \stackrel{\to }{AB}×\stackrel{\to }{AC}=\left[\begin{array}{ccc}i& j& k\\ -1& 2& 2\\ 1& 1& -1\end{array}\right]\\ =\left(-2-2\right)i-\left(1-2\right)j+\left(-1-2\right)k\\ =-4i+j-3k\\ A=\frac{1}{2}\left(\sqrt{16+1+9}\right)\\ =\frac{1}{2}\left(\sqrt{26}\right)\\ =\frac{\sqrt{26}}{2}\\ \text{(b) area of}\mathrm{△}ABC=\frac{1}{2}||\stackrel{\to }{AB}||h\\ h=\frac{\text{area of}\mathrm{△}ABC}{\frac{1}{2}||\stackrel{\to }{AB}||}\\ =\frac{\frac{\sqrt{26}}{2}}{\frac{1}{2}\sqrt{1+4+4}}\\ =\frac{\frac{\sqrt{26}}{2}}{\frac{1}{2}\sqrt{9}}\\ =\frac{\frac{\sqrt{26}}{2}}{\frac{3}{2}}\\ =\frac{\sqrt{26}}{3}\end{array}$