Post by: Husky on June 09, 2020, 03:39:29 pm
Relative to a fixed origin, the points A, B and C are defined respectively by the position vectors (a = -i - j), (b = 3i +2j) and (c = -m + 2j), where m is a real constant.
If the magnitude of angle ABC is 60 degrees (pi/3), find m.
Thanks in advance for any help.
Post by: S_R_K on June 09, 2020, 05:37:17 pm
The question:
Relative to a fixed origin, the points A, B and C are defined respectively by the position vectors (a = -i - j), (b = 3i +2j) and (c = -m + 2j), where m is a real constant.
If the magnitude of angle ABC is 60 degrees (pi/3), find m.
Thanks in advance for any help.
Use
Post by: Husky on June 09, 2020, 09:40:57 pm
UseI tried that and got m = -3 but I think it’s wrong cause m can’t equal -3 (denominator becomes zero).and solve for m.
Post by: RuiAce on June 09, 2020, 11:17:04 pm
I'm assuming that where you wrote \( \mathbf{c} =-\mathbf{m}+2\mathbf{j}\) where \(\mathbf{m}\) is a real constant, that you actually meant \(\mathbf{c}=-m\mathbf{i}+2\mathbf{j}\), where \(m\) is a real constant. But then the points \(B\) and \(C\) corresponding to the position vectors \(\mathbf{b}\) and \(\mathbf{c}\) have the same \(\mathbf{j}\) component, so the point \(C\) is always gonna be a fixed point to the left or the right of \(B\). As a consequence, \(\angle ABC\) will already have a pre-determined value, and apparently it's not pi/3 either.
Post by: wsdm on June 09, 2020, 11:20:43 pm
Assuming we're working with \(\angle{ABC}\), then you have \(\overrightarrow{OB} = 3i+2j\) and \(\overrightarrow{OC} = -mi+2j\), right? However, they would have the same \(y\)-coordinate placement (see attachment below), so then technically the maximum angle between vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) would have to be \(\tan\theta = \frac{3}{4}\) (approximately \(\theta = 36.87^{\circ}\)).
Post by: wsdm on June 09, 2020, 11:22:33 pm
Are you absolutely sure there's no typo in your question?Oops, looks like I was a minute too late in replying.