 # I need help with Linear Algebra Engineering Anonymous
timer Asked: Nov 12th, 2015
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### Question Description alancheng
School: UIUC   Thank you for the opportunity to help you with your question!

The AREA of triangle ABC is (1/2) * square root of 45  square units. which equals  3.3541 square units.

We need to find vectors AB and AC.

Our Area formula is  Area = (1/2)* || AB || * ||AC|| * sin w.

for w is the angle between vectors AB and AC.

There is an equivalent formula.  The cross product.  where || AB  x AC || =|| AB || * ||AC|| * sin w

AB = <4 - 3, -2 - (-1), 6 - 4 > = < 1, -1, 2>

AC = <5 - 3, 0-(-1), 2-4> = <2, 1, -2>

||AB x AC|| =  magnitude of determinant of   matrix:

<i, j, k>

<1, -1, 2>

<2, 1, -2>

determinant = ( (-1)*(-2) - 1*2) i -  (-2 - 4)*j  + (1 - (-2))k

determinant = <0, 6, 3>

the magnitude is  square root of (6*6 + 3*3) =  square root 45.

Area = (1/2)* sqrt(45) = 3.3541 sq. units

Please let me know if you need any clarification. I'm always happy to answer your questions.

flag Report DMCA  Review Anonymous
Good stuff. Would use again. Brown University

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