find the area of the parallelogram. A(1,1,3) B(−7,6,11) C(−5,9,4) D(3,4,−4)

Question

I keep getting 7(113)^½ as my answer, but my online homework says this is incorrect

Answer 1

Let a=AB=CD=√((-7-1)²+(6-1)²+(11-3)²)=

√((3-(-5))²+(4-9)²+(-4-4)²)=

√(8²+5²+8²)=√153=3√17. 

Let b=BC=AD=√((-5-(-7))²+(9-6)²+(4-11)²=

√((3-1)²+(4-1)²+(-4-3)²)=

√(4+9+49)=√62.

a and b are the lengths of two sides of the parallelogram. The area of the parallelogram is absinθ where θ is the angle between the two sides.

The dot product of vectors AB and BC=(1)(-7)+(1)(6)+(3)(11)=

-7+6+33=32=abcosθ. 

So (abcosθ)²=1024. (abcosθ)²+(absinθ)²=(ab)²=153×62=9486.

(absinθ)²=9486-1024=8462. Therefore the area of the parallelogram is √8462=91.9891 sq units approx.