3D Triangles (VERTICES), and Vectors (Help)

[HugeLag]

Hello, I am working on quite a tough question. I have been practicing and practicing my 3D plotting skills, as well as watching online videos on how to draw in a 3D space it. It just seems to be coming out wrong, and this question is driving me mad. Could someone have a look at this, and tell me if there is some sort of method to work this out without drawing it all out? I understand the xyz coordinate system, its just the plotting part, and finding the perpendicular vector.

Question 17) The vertices of a triangle in 3D space are A(2.0, 1.0, 0.0), B(4.7, 3.0, 2.0), C(4.7, 3.1, -1.0). Which of the following vectors perpendicular to the triangle?

Vector 1) (2.5, -3.2, 0.22)

Vector 2) (-3.2, -7.1, 1.3)

Vector 3) (-6.2, 8.1, 0.27)

Vector 4) (3.2, 6.5, 0.17)

All help, and working methods are greatly appreciated!

 

[stapel]

Question 17) The vertices of a triangle in 3D space are A(2.0, 1.0, 0.0), B(4.7, 3.0, 2.0), C(4.7, 3.1, *1.0). Which of the following vectors perpendicular to the triangle?

Vector 1) (2.5, *3.2, 0.22)

Vector 2) (*3.2, *7.1, 1.3)

Vector 3) (*6.2, 8.1, 0.27)

Vector 4) (3.2, 6.5, 0.17)

What mathematical meaning has your book or instructor given to the asterisks in front of some of the coordinates? Thank you!

 

[HugeLag]

What mathematical meaning has your book or instructor given to the asterisks in front of some of the coordinates? Thank you!

sorry

I dont know why but when I paste my question and it gets moderated asterisks appear. They are actual minuses I amended the original post. Sorry for the confusion.

 

[pka]

Question 17) The vertices of a triangle in 3D space are A(2.0, 1.0, 0.0), B(4.7, 3.0, 2.0), C(4.7, 3.1, -1.0). Which of the following vectors perpendicular to the triangle?
Vector 1) (2.5, -3.2, 0.22)
Vector 2) (-3.2, -7.1, 1.3)
Vector 3) (-6.2, 8.1, 0.27)
Vector 4) (3.2, 6.5, 0.17)

I still hate your courses' notation.
\(\displaystyle \overrightarrow {AB} = \left\langle {2.7,2,2} \right\rangle ~\&~ \overrightarrow {AC} = \left\langle {2.7,2.1, - 1} \right\rangle\)

Your answer is \(\displaystyle \left\langle {2.7,2,2} \right\rangle \times \left\langle {2.7,2.1, - 1} \right\rangle\)

 

[HugeLag]

I still hate your courses' notation.
\(\displaystyle \overrightarrow {AB} = \left\langle {2.7,2,2} \right\rangle ~\&~ \overrightarrow {AC} = \left\langle {2.7,2.1, - 1} \right\rangle\)

Your answer is \(\displaystyle \left\langle {2.7,2,2} \right\rangle \times \left\langle {2.7,2.1, - 1} \right\rangle\)

ThankYou!

Just to be sure. When I worked out the cross-product my answer was the following. (-6.2,8.1,0.27). Please can you confirm that this is the correct answer.

Also sorry about the notations, I guess that is the way they teach it at uni?

 

[pka]

ThankYou!

Just to be sure. When I worked out the cross-product my answer was the following. (-6.2,8.1,0.27). Please can you confirm that this is the correct answer.

It is correct.