Vectors Using Cosine or Sin Law

[chee]

Need help with a question ... .__."

q: if the magnitude of a=2 and the magnitude of 5a-2b= 7.7, and the angles between the vectors a and b is 50 degrees determine the magnitude of b.

first I drew a diagram... then i used cosine law to solve for b

l5a-2bl = (5x2) + (2b)^2 -2(5x2)(2b)cos50

7.7= 10 +4b^2 -40b cos 50

this worked out to be a quadratic, the zeros i found were 6.3 & 0.09, but then i subbed those numbers back into the cosine law i don't get back 7.7...

I tried this question like this, l5a-2bl = (5x2)^2 + (2b)^2 -2(5x2)(2b)cos50, but it didn't work out either.

 

[pka]

Need help with a question ... .__."
q: if the magnitude of a=2 and the magnitude of 5a-2b= 7.7, and the angles between the vectors a and b is 50 degrees determine the magnitude of b.

From the given we know:
\(\displaystyle \|a\|=2\) , \(\displaystyle 25\|a\|^2-20(a\cdot b)+4\|b\|^2=(7.7)^2\) and \(\displaystyle cos(50^o)=\dfrac{a\cdot b}{\|a\|\|b|}\).

From all that we get a quadratic in \(\displaystyle \|b\|\)

 

[pappus]

Need help with a question ... .__."

q: if the magnitude of a=2 and the magnitude of 5a-2b= 7.7, and the angles between the vectors a and b is 50 degrees determine the magnitude of b.

first I drew a diagram... then i used cosine law to solve for b

l5a-2bl = (5x2) + (2b)^2 -2(5x2)(2b)cos50

7.7= 10 +4b^2 -40b cos 50

this worked out to be a quadratic, the zeros i found were 6.3 & 0.09, but then i subbed those numbers back into the cosine law i don't get back 7.7...

I tried this question like this, l5a-2bl = (5x2)^2 + (2b)^2 -2(5x2)(2b)cos50, but it didn't work out either.

1. Please don't x as the multiplication sign, better use an asterix *

2. This equation is wrong:

l5a-2bl = (5x2) + (2b)^2 -2(5x2)(2b)cos50

This should be:

\(\displaystyle (5a-2b)^2 = (5 \cdot 2)^2 + (2b)^2 -5 \cdot (5 \cdot 2) \cdot 2b \cdot \cos(50^\circ)\)

By the way : \(\displaystyle \cos(50^\circ) \ne \cos(50)\)