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 November 5th, 2014, 10:14 AM #1 Newbie   Joined: Oct 2014 From: Marilia Posts: 6 Thanks: 0 find area of triangle (linear algebra ) Guys, good afternoon! I'm racking my brain with this: a) Obtain the area of ​​the triangle vertices A ( 1,0,1 ) B ( 0,2,3 ) and C ( 2,0,1 ) b ) Use the result of the area to FIND the height of the vertex C to the side AB. Can anyone enlighten me with making the resolution of this exercise? Last edited by skipjack; November 5th, 2014 at 06:05 PM.
 November 5th, 2014, 10:24 AM #2 Newbie   Joined: Nov 2014 From: mississauga Posts: 2 Thanks: 0 Could you create 2 vectors with those 3 coordinates, then use 1/2 the magnitude of the cross product? Last edited by skipjack; November 5th, 2014 at 06:05 PM.
November 5th, 2014, 12:38 PM   #3
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Imitating $\displaystyle \color{green}{\textbf{soroban}}$'s way of posting.

Hello, Juan Victor!
Quote:
 a) Obtain the area of ​​the triangle vertices A ( 1,0,1 ) B ( 0,2,3 ) and C ( 2,0,1 )
We have : $\displaystyle A \ \left( \ 1,0,1 \ \right) \ B \ \left( \ 0,2,3 \ \right) \text{ and } C \ \left( \ 2,0,1 \ \right)$.

The magnitude of the cross product of two vectors is the area of the parallelogram which has adjacent sides equal to the magnitudes of the two vectors.

We form the vectors $\displaystyle \overline{\text{AB}}$ and $\displaystyle \overline{\text{AC}}$

$\displaystyle \overline{\text{AB}}= \bigg( \ 0-1 \ , \ 2-0 \ , 3-1 \bigg)= \bigg( \ -1 \ , \ 2 \ , \ 2 \ \bigg)$

$\displaystyle \overline{\text{AC}}= \bigg( \ 2-1 \ , \ 0-0 \ , 1-1 \bigg)= \bigg( \ 1 \ , \ 0 \ , \ 0 \ \bigg)$

We find their cross product, i.e., $\displaystyle \overline{\text{AB}} \times \overline{\text{AC}}$

$\displaystyle \begin{array}{| r c c |} \ i & j & k \ \\ -1 & 2 & 2 \ \\ \ 1 & 0 & 0 \ \end{array}= \Big[ \ (2)(0)-(2)(0) \ \Big] \overline{\text{i}} \ - \ \Big[ \ (-1)(0)-(2)(1) \ \Big] \overline{\text{j}} \ + \ \Big[ (-1)(0)-(2)(1) \ \Big] \overline{\text{k}}$

$\displaystyle = 0 \overline{\text{i}} \ - \ (0-2) \overline{\text{j}} \ + \ (-1-2) \overline{\text{k}}$

$\displaystyle = \bigg( \ 0 \ , \ 2 \ , -3 \ \bigg)$

The magnitude of $\displaystyle = \bigg( \ 0 \ , \ 2 \ , -3 \ \bigg)$ is
$\displaystyle \sqrt{0^{2}+2^{2}+(-3)^{2}}= \sqrt{0+4+9}= \sqrt{13}$

The triangle's area is $\displaystyle \frac{1}{2}$ the area of the parallelogram. Can you proceed? Can you solve the second part of the question?

Last edited by skipjack; November 5th, 2014 at 06:12 PM.

 November 5th, 2014, 06:09 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,722 Thanks: 1807 The method used above is okay, but there's a slip in the working. Alternatively, use the easily found "height" from vertex B to side AC and the fact that AC has length 1.
 November 6th, 2014, 01:15 AM #5 Newbie   Joined: Oct 2014 From: Marilia Posts: 6 Thanks: 0 Yes, Thanks Prakhar!
 November 6th, 2014, 01:32 AM #6 Newbie   Joined: Oct 2014 From: Marilia Posts: 6 Thanks: 0 Prakrar only one observation, [(-1 )(0 ) - (2 )(1)] k = 0i - (0-2)j + ( -1-2 )k . = ( 0, 2, -3) The value of the last element of the vector k is related to the -2 and -3 do not agree?
 November 6th, 2014, 06:10 AM #7 Global Moderator   Joined: Dec 2006 Posts: 19,722 Thanks: 1807 You've found prakhar's slip... the last line should be = (0, 2, -2).

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