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

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up