perpendicular vectors

[akoaysigod]

This is a problem from a physics lab and I'm not quite sure where to begin because its been a while since I've done this.

There are 3 vectors of magnitude 100, 150, 200. The vector of length 100 is at 0 theta. I'm trying to find of which length and theta value would make one of these perpendicular. Or so that it adds to 0. I'm not sure how exactly to word it. So one vector cancels out the other two vectors I guess maybe is the best way to put it.

Due to time constraints we just did this visually and we got the values of roughly 286.5 degrees. The one perpendicular is at roughly 134 degrees. Which is relatively close. I'm posting this in the calculus section because I know we did this as part of the intro to calc 3. But the angles are throwing me off.

<100, 0, 0> dot <x, y, z> = 0 but that's all I've got and this is technically in 2D. Any hints on where to start or what direction I should go in with this problem would be great.

Thanks

 

[Deleted member 4993]

This is a problem from a physics lab and I'm not quite sure where to begin because its been a while since I've done this.

There are 3 vectors of magnitude 100, 150, 200. The vector of length 100 is at 0 theta. I'm trying to find of which length and theta value would make one of these perpendicular. Or so that it adds to 0. I'm not sure how exactly to word it. So one vector cancels out the other two vectors I guess maybe is the best way to put it.

Due to time constraints we just did this visually and we got the values of roughly 286.5 degrees. The one perpendicular is at roughly 134 degrees. Which is relatively close. I'm posting this in the calculus section because I know we did this as part of the intro to calc 3. But the angles are throwing me off.

<100, 0, 0> dot <x, y, z> = 0 but that's all I've got and this is technically in 2D. Any hints on where to start or what direction I should go in with this problem would be great.

Thanks

suppose the other two vectors are 200(ai + bj) and 150(ci + dj)

then

200*b + 150 * d = 0

200*a + 150*c = -100

and

a[sup:16cils8r]2[/sup:16cils8r] + b[sup:16cils8r]2[/sup:16cils8r] = 1

c[sup:16cils8r]2[/sup:16cils8r] + d[sup:16cils8r]2[/sup:16cils8r] = 1

 

[soroban]

Hello, akoaysigod!

Your description is very fuzzy . . .

This is a problem from a physics lab.

There are 3 vectors of magnitude 100, 150, 200.
The vector of length 100 is at 0 theta.
I'm trying to find of which length and theta value would make one of these perpendicular, or so that it adds to 0.
I'm not sure how exactly to word it.
So one vector cancels out the other two vectors . . . I guess maybe is the best way to put it.

If you mean "the sum of the three vectors is zero", there is a solution.

    C
    o - - - - - - F
     *  *
      *     *
       *        *    200
    150 *           *
         *              *
          *                 *
            *                    *
    E - - - o  *  *  *  *  *  *  *  o - - - D 
            A         100           B

\(\displaystyle \text{I will write the vectors in }\langle r,\,\theta\rangle\text{ form.}\)

\(\displaystyle \text{So: }\:\boxed{\overrightarrow{AB} \:=\:\langle 100,\,0^o\rangle}\)


\(\displaystyle \text{The sum of three vectors is zero if they form a triangle ("tip to tail").}\)


\(\displaystyle \text{In }\Delta ABC:\)

\(\displaystyle \text{Law of Cosines: }\;\cos A \;=\;\frac{100^2 +150^2 - 200^2}{2(100)(150)} \;=\;-0.25 \quad\Rightarrow\quad A \:\approx\:104.5^o\)

\(\displaystyle \text{Law of Cosines: }\;\cos C \;=\;\frac{150^2 + 200^2 - 100^2}{2(150)(200)} \;=\;0.875 \quad\Rightarrow\quad C \;\approx\;29.0^o\)

\(\displaystyle \text{Then: }\;\angle B \;=\;180^o - 104.5^o - 29.0^i \;=\;46.5^o \quad\Rightarrow\quad \angle CBD \:=\:133.5^o\)

. . \(\displaystyle \text{Therefore: }\;\boxed{\overrightarrow{BC} \;=\;\langle 200,\:133.5^o\rangle}\)


\(\displaystyle \angle CAE \;=\;180^o - 104.5^o \;=\;75.5^o \;=\;\angle FCA\)

. . \(\displaystyle \text{Therefore: }\;\boxed{\overrightarrow{CA} \;=\;\langle 150,\:-75.5^o\rangle}\)