1. cartesian coordinates

1. Points A, B and C are the three vertices of a triangle in 3D space (Cartesian coordinate system). It is known that
AB = (3,6,1) BC = (0,-7,-4) CA = (-3,1,3)
[Note: A right-angled triangle is a triangle with one 90o internal angle.]

The Cartesian coordinates of Point A are (3, 4, 1).
(b) What are the Cartesian coordinates of Point C?
(c) Find a vector that is perpendicular to the plane that contains triangle ABC.

Another point, Point D, is in the same 3D space such that the vector is perpendicular to the plane that contains triangle ABC.

(d) Given that AD = (34, a, b), determine the values of a and b.

2. Hello, dorwei92!

I must assume that this is a vector problem.

Here's the first part . . .

1. Points $\displaystyle A, B, C$ are the three vertices of a triangle in 3-space.

It is known that: .$\displaystyle AB = \langle3,6,1\rangle,\;\;BC = \langle0,-7,-4\rangle,\;\;CA = \langle-3,1,3\rangle$

(a) Is $\displaystyle \Delta ABC$ a right triangle? Justify your answer.
Consider the lengths (magnitudes) of the three vectors:

. . $\displaystyle \begin{array}{ccccc}|AB| &=& \sqrt{3^2+6^2+1^2} &=& \sqrt{46} \\ |BC| &=& \sqrt{0^2 + (\text{-}7)^2 + (\text{-}4)^2} &=& \sqrt{65} \\ |CA| &=& \sqrt{(\text{-}3)^2 + 1^2 + 3^2} &=& \sqrt{19} \end{array}$

We find that: .$\displaystyle AB^2 + CA^2 \:=\:BC^2$

. . That is: .$\displaystyle (\sqrt{46})^2 + (\sqrt{19})^2 \;=\;(\sqrt{65})^2 \quad\Rightarrow\quad 46 + 19 \:=\:65$

Hence, the three sides satisfy the Pythagorean Theorem

Therefore, $\displaystyle \Delta ABC$ is a right triangle.