Magnitude and direction of a resultant vector.

**Pupil** · May 29th 2011, 10:26 PM

Problem: Find the magnitude and direction of the resultant of two vectors $\vec{u}$ and $\vec{v}$ with magnitudes of 6 and 8 and an angle of 60° between them.

Okay, so the first thing I do is find θ by subtracting 60° from 180° and yield 120°. Then I use cosine law to find the magnitude:
$| \vec{u}+\vec{v} |^{2} = | \vec{u} |{^2}+| \vec{v} |^{2} - 2 | \vec{u} | | \vec{v} | cos\theta$
$| \vec{u}+\vec{v} |^{2} = (6)^{2} + (8)^{2} - 2|6||8|cos(120°)$
$| \vec{u}+\vec{v} |^{2} = \sqrt{148}$

Alright, so I got the magnitude right according to my book, and then I go on to draw the vectors, tail-to-tail and then the resultant: Vector diagram

So, now I use sine law to find angle alpha:
$\frac{sin\alpha}{|\vec{u}|} = \frac{sin120°}{|\vec{u}+\vec{v}|}$
$\frac{\sin\alpha }{6} = \frac{\sin120°}{\sqrt{148}}$
$\sin\alpha =\frac{(6)\sin120°}{\sqrt{148}}$
$\sin\alpha \approx 0.4271$
$\alpha \approx \sin^{-1}(0.4271) \approx 25.285$ ° is the direction I get after using sine law, but the answer is completely different in my textbook. The reason why is because they drew $\vec{u}$ first then $\vec{v}$ tail to tail, but I did not learn any law in which there is a general convention to draw vectors of differing magnitudes from tail to tail, therefore I dismissed it as arbitrary. If there is no law, then by using sine law, shouldn't I have arrived at the same direction as the book? The answer in the book:
α ≈ 34.7°, "the resultant of $\vec{u}$ and $\vec{v}$ is approximately 34.7° from $\vec{u}$ ."

So, where did I go wrong? Thanks in advance.

**HallsofIvy** · May 30th 2011, 12:17 AM

You didn't go wrong. However you should be able to see from your diagram that you calculated the angle the resultant makes with $\vec{v}$ , not $\vec{u}$ . You answer is correct if you say which vector it is relative to.

Note that 34.7+ 28.3= 60.

**Pupil** · May 30th 2011, 12:21 AM

Originally Posted by HallsofIvy

You didn't go wrong. However you should be able to see from your diagram that you calculated the angle the resultant makes with $\vec{v}$ , not $\vec{u}$ . You answer is correct if you say which vector it is relative to.

Note that 34.7+ 28.3= 60.

Ah, I see, thank you HallsofIvy.

Just one more question: Let's say the angle between the vectors is greater than 180°. Would it then be subtracted from 360° or how would I generally find angle theta before using cosine law? Even though I haven't dealt with questions that ask for vectors with angles greater than 180° between them, I'd still like to know. Thanks in advance.

**HallsofIvy** · May 30th 2011, 03:09 AM

The angle between two vectors cannot be larger than 180 degrees!

**Pupil** · May 31st 2011, 11:26 AM

Originally Posted by HallsofIvy

The angle between two vectors cannot be larger than 180 degrees!

How come? Would they no longer be considered be considered vectors if the angle is greater than 180 degrees? Is there a specific law that states this or is it common mathematical knowledge? I'd like to write this in my notes, thank in advance HallsofIvy.

**HallsofIvy** · June 1st 2011, 04:50 AM

Given two vectors there are always two ways to measure the angle between them- clockwise and counter clockwise. And, unless the vectors are pointing in exactly opposite directions (one is a negative number times the other), one angle is less than 180 degrees and the other larger than 180 degrees. "The" angle between two vectors is defined to be the smaller of those two angles.

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