Question on Vectors involving Triangle

[Edder]

Hey everyone, I am having a bit of difficulty with a problem involving vectors. It says:

A triangle ∆PQR in 3-space has vertices: P(2,4,0), Q(−3,5,3), R(−2,2,1). Use vectors to decide which one of the following properties the triangle has.

How exactly would I solve this problem? I first tried taking various dot products to find if any of them equaled to zero. I then found vectors through the line segments of PQ, QR, and RP, and I still haven't quite figured out what to do with it. The question asks at what vertices there is s right angle. Any thoughts?

Any feedback is appreciated guys, thanks.

 

[soroban]

Hello, Edder!

You didn't show us your work.
How can we point out your error?

A triangle PQR in 3-space has vertices: P(2,4,0), Q(−3,5,3), R(−2,2,1).
Use vectors to decide which one of the following properties the triangle has.

\(\displaystyle \text{The vectors are: }\:\begin{Bmatrix} \overrightarrow{PQ} &=& \langle \text{-}5,1,3\rangle \\ \overrightarrow{QR} &=& \langle 1,\text{-}3,\text{-}2\rangle \\ \overrightarrow{PR} &=& \langle \text{-}4,\text{-}2,1\rangle \end{Bmatrix}\)

\(\displaystyle \text{We see that: }\:\overrightarrow{QR}\cdot\overrightarrow{PR} \:=\:\langle 1,\text{-}3,\text{-}2\rangle\cdot\langle \text{-}4,\text{-}2,1\rangle \:=\:\text{-}4 + 6 -2 \:=\:0\)

\(\displaystyle \text{Hence: }\:\overrightarrow{QR} \perp \overrightarrow{PR}\)

We have a right triangle with /R = 90^o and hypotenuse \(\displaystyle PQ.\)

 

[Deleted member 4993]

Hey everyone, I am having a bit of difficulty with a problem involving vectors. It says:

A triangle ∆PQR in 3-space has vertices: P(2,4,0), Q(−3,5,3), R(−2,2,1). Use vectors to decide which one of the following properties the triangle has.

How exactly would I solve this problem? I first tried taking various dot products to find if any of them equaled to zero. I then found vectors through the line segments of PQ, QR, and RP, and I still haven't quite figured out what to do with it. The question asks at what vertices there is s right angle. Any thoughts?

Any feedback is appreciated guys, thanks.

Which properties are you referring here?

 

[Edder]

Hello, Edder!

You didn't show us your work.
How can we point out your error?


\(\displaystyle \text{The vectors are: }\:\begin{Bmatrix} \overrightarrow{PQ} &=& \langle \text{-}5,1,3\rangle \\ \overrightarrow{QR} &=& \langle 1,\text{-}3,\text{-}2\rangle \\ \overrightarrow{PR} &=& \langle \text{-}4,\text{-}2,1\rangle \end{Bmatrix}\)

\(\displaystyle \text{We see that: }\:\overrightarrow{QR}\cdot\overrightarrow{PR} \:=\:\langle 1,\text{-}3,\text{-}2\rangle\cdot\langle \text{-}4,\text{-}2,1\rangle \:=\:\text{-}4 + 6 -2 \:=\:0\)

\(\displaystyle \text{Hence: }\:\overrightarrow{QR} \perp \overrightarrow{PR}\)

We have a right triangle with /R = 90^o and hypotenuse \(\displaystyle PQ.\)

Thank you for that thorough explanation, I appreciate it. I made an error on my line segments. I added the vertices of PQ instead of subtracting like you did. That is why I couldn't get my dot products to equal zero.

Which properties are you referring here?

At which vertices there is a right angle.

 

[Edder]

Also, when combining vertices to find the vectors, why do you subtract them? I thought that you would simply add them together.

 

[Yogi]

Take it down to 1 dimension, think of the number line. Start at +4, move to +5. Did you move +9 or +1? Add or subtract? The same idea applies to 3 dimensions.

 

[pappus]

Also, when combining vertices to find the vectors, why do you subtract them? I thought that you would simply add them together.

1. Draw a sketch (see attachment)

2. Use your index finger: If you want to "go" from Q (with the staionary vector \(\displaystyle \vec q\)) to R (with the staionary vector \(\displaystyle \vec r\)) your finger will move \(\displaystyle \vec q\) backwards that means it is \(\displaystyle - \vec q\), and afterwards your finger will move along \(\displaystyle \vec r\).

3. That means \(\displaystyle \overrightarrow{QR} = - \vec q + \vec r = \vec r - \vec q\). You certainly have noticed that the letters are ordered symmetrically to the equal sign:

\(\displaystyle \overrightarrow{QR} = \vec r - \vec q\)

 

[Edder]

Got it. Thanks.