# Thread: area of triangle using 3d vectors

1. ## area of triangle using 3d vectors

can someone explain how to do this problem, i got the first part correct but i dont understand how to get the area exactly. THanks!
Consider the points below. P(1, 0, 0), Q(0, -2, 0), R(0, 0, -3)

(a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.
Enter a number.
6i + Enter a number.
-3j + Enter a number.
-2k

(b) Find the area of the triangle PQR.
Enter an exact number as an integer fraction or decimal.
3

NVM I FIGURED IT OUT THANKS FOR LOOKING AT IT!!

2. Originally Posted by ahawk1
can someone explain how to do this problem, i got the first part correct but i dont understand how to get the area exactly. THanks!
Consider the points below. P(1, 0, 0), Q(0, -2, 0), R(0, 0, -3)

(a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.
Enter a number.
6i + Enter a number.
-3j + Enter a number.
-2k

(b) Find the area of the triangle PQR.
Enter an exact number as an integer fraction or decimal.
3
It will be half of the magnitude of the cross product of PQ and RQ
$PQ=\vec i +2 \vec j$ and $RQ =2 \vec j - 3\vec k$

$\begin{vmatrix}
i & j & k \\
1 & 2 & 0 \\
0 & 2 & -3 \\
\end{vmatrix} = (6-0)\vec i-(-3-0)\vec j+(2-0)\vec k=6 \vec i +3 \vec j +2\vec k$

Now we take half of the magnitude to get

$\frac{1}{2}\sqrt{(6)^2+(3)^2+(2)^2}=\frac{1}{2}\sq rt{49}=\frac{7}{2}$