Thank you for the opportunity to help you with your question!

Given th4e vertices of the triangle

A(3,-1,4) ,B(4,-2,6) and C(5,0,2)

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

BC=><5-4,0-(-2),2-6> => <1,2,-4>

Now the magnitude of AB,|AB|=

sqrt(1^2+1^2+2^2)=sqrt6

The angle between the two vector ,

cos theta =[AB.BC]/|AB||BC|=-.8017

sin(theta)=.3572

Therefore the area of the triangle is =

A =.5*sqrt6*sqrt21*.5976 =3.5(answer)

**Mistyped"So,sin(theta)=.3572"

Please make sure to see that **sin(theta)=.5766 and not sin(theta)=.3572(mistyped)

Sure this right.

Yes 100% I have also checked my answer using calculator.Just make the correction the area is 3.35.I did a mistake in multiplying at the last

.5*sqrt6*sqrt21*.5976=3.35**

I am also sending the snapshot of the calculator answer.

thanks

could u write how u solved it on a sheet of paper then send it?

Hey,Why you gave me a negative review although I answered correctly.You would have asked me your doubt and then decide whether my explanation was effective or not.

Thanks

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