Finding an area of a triangle formed by three points

• CrosisBH
In summary, the conversation discusses finding an orthogonal vector to a plane through three given points, and calculating the area of the triangle formed by those points. The solution involves finding the cross product of two vectors and taking the magnitude of the resulting vector. In the end, the correct answer is found to be 13√5/2.
CrosisBH

Homework Statement

P(3, 0, 3), Q(−2, 1, 5), R(6, 2, 7)
(a) Find a nonzero vector orthogonal to the plane through the points P, Q,
and R.

(b) Find the area of the triangle PQR

Homework Equations

$$A = \frac{1}{2}|\vec{AB}\times\vec{AC}|$$
Source: http://www.maths.usyd.edu.au/u/MOW/vectors/vectors-11/v-11-7.html

The Attempt at a Solution

The first part was straight forward and my homework marked it right:

$$P(3,0,3)\\ Q(-2,1,5)\\ R(6,2,7)\\ \vec{PR} = \langle 3,2,4 \rangle\\ \vec{PQ} = \langle 5,1,2 \rangle$$
To find a vector orthogonal to two vectors, you do a cross product:
$$\vec{PR} \times \vec {PQ} = \langle 0,14,-7 \rangle$$
My homework marked this part right for this answer.

Part two:
First taking the magnitude of the cross product:
$$|\vec{PR} \times \vec {PQ}| = \sqrt{245} = 7\sqrt{5}$$
Dividing it by two:
$$A = \frac{7}{2}\sqrt{5}$$
Which is my final answer, and the website marks it as wrong, but I can't see why it's wrong. The magnitude is right which I triple checked, by my own hands and Wolfram Alpha, and the order of the cross product is irrelevant in this situation.

I found a calculator that calculates the area from three 3D points and it spit out the nonexact answer of 14.534. My answer is approximately 7.826, a little more than half. I can't see my error unless I'm completely calculating this area wrong. Thanks for the help.

Attachments

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##\vec {PQ}## looks wrong.

fresh_42 said:
##\vec {PQ}## looks wrong.
AHHH it was that simple. I was successful in finding a orthogonal vector, but it wasn't the cross product of the two supposed vectors from the points.

Correction:
$$\vec{PQ} = \langle -5, 1, 2 \rangle$$
Making the normal vector of interest
$$\vec{PR} \times \vec {PQ} = \langle 0,-26,13 \rangle$$
$$|\vec{PR} \times \vec {PQ}| = 13\sqrt{5}$$
$$\frac{13\sqrt{5}}{2}$$

Let it be the simple subtraction that kills me haha. Thank you

1. How do I find the area of a triangle formed by three points?

To find the area of a triangle formed by three points, you can use the formula A = 1/2 * base * height, where the base is the distance between any two of the points, and the height is the perpendicular distance from the third point to the base.

2. Can I use the distance formula to find the base and height?

Yes, you can use the distance formula, d = √((x2-x1)^2 + (y2-y1)^2), to find the base and height of the triangle. Simply plug in the coordinates of two of the points to find the distance between them, and then use the perpendicular distance formula to find the height.

3. What if the three points are collinear?

If the three points are collinear, meaning they lie on the same line, the area of the triangle formed by them will be 0. This is because the base and height will also be 0, making the area calculation 1/2 * 0 * 0 = 0.

4. Can I use the shoelace formula to find the area?

Yes, the shoelace formula, also known as the surveyor's formula, can also be used to find the area of a triangle formed by three points. This method involves finding the sum of the products of the coordinates of the points, and then taking half of that value.

5. Are there any other methods for finding the area of a triangle formed by three points?

Yes, there are other methods such as using determinants or vectors. However, the most commonly used and simplest method is using the formula A = 1/2 * base * height. It is important to choose a method that you are comfortable with and can easily understand and apply.

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