# I need help with Linear Algebra Engineering

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### Question Description

alancheng
School: UIUC

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

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Review

Anonymous
Good stuff. Would use again.

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