If u and v are both unit vectors that make an angle of 60 degrees to each other, calculate
a) |2u+3v|
b) |4u-5v|
The answer key says the correct answer for a) is 4.71 and the correct answer for b) is 3.22.
a)|2u+3v|^2=|2|^2+|3|^2-2|2||3|cos(60)=7; 7^.5=~2.65
b|4u-5v|^2=|4|^2+|5|^2-2|4||5|cos(60)=21; 21^.5=~4.58
What am I doing wrong?
You should draw the parallelograms. For (a), you found the short diagonal. What you should have used was the 120° angle formed by placing the start of 3v at the tip of 2u. That means that you should have used
|2|^2+|3|^2-2|2||3|cos(120)=13+6=19; √19=4.36
4.71 is wrong
To check, if you add the components, you get
w=2u+3v=(2,0)+(3cos60,3sin60)=(2,0)+(1.5,1.5√3)
take the length of that and you get |w|^2 = 3.5^2 + 2.25*3 = 19
Do a similar recalculation for part (b). Maybe they key is wrong there as well.
Your calculations for the magnitudes of |2u+3v| and |4u-5v| are incorrect. Let's go through the correct calculations:
a) To find |2u+3v|, we can use the formula for the magnitude of a vector:
|2u+3v| = √((2u+3v) · (2u+3v)), where "·" denotes the dot product.
Now, let's substitute the values:
|2u+3v| = √((2u) · (2u) + (2u) · (3v) + (3v) · (2u) + (3v) · (3v))
Since u and v are both unit vectors, their magnitudes are 1. Also, the dot product of two vectors making an angle of 60 degrees is equal to half the product of their magnitudes. Therefore, we have:
|2u+3v| = √((2)^2 + (2)(3)(1/2) + (3)(2)(1/2) + (3)^2)
= √(4 + 6 + 6 + 9)
= √25
= 5
So, the correct magnitude of |2u+3v| is 5, not 4.71.
b) Similarly, to find |4u-5v|, we can use the same formula:
|4u-5v| = √((4u-5v) · (4u-5v))
Substituting the values:
|4u-5v| = √((4u) · (4u) + (4u) · (-5v) + (-5v) · (4u) + (-5v) · (-5v))
Since u and v are both unit vectors, their magnitudes are 1. Also, the dot product of two vectors making an angle of 60 degrees is equal to half the product of their magnitudes. Therefore, we have:
|4u-5v| = √((4)^2 + (4)(-5)(1/2) + (-5)(4)(1/2) + (-5)^2)
= √(16 - 10 - 10 + 25)
= √21
≈ 4.58
So, the correct magnitude of |4u-5v| is approximately 4.58, not 3.22.
In your calculations, you squared the magnitudes of the coefficients (2, 3, 4, 5) instead of squaring the entire vector resulting from multiplying each coefficient with the respective unit vector. To correct your calculations, follow these steps:
a) |2u+3v|^2 = (2u + 3v) · (2u + 3v)
= (2u) · (2u) + (2u) · (3v) + (3v) · (2u) + (3v) · (3v)
= 4(u · u) + 6(u · v) + 6(v · u) + 9(v · v)
= 4(|u|^2) + 6(u · v) + 6(v · u) + 9(|v|^2)
= 4(1) + 6(u · v) + 6(v · u) + 9(1)
= 4 + 6(1)(cos(60)) + 6(cos(60)) + 9
Using cos(60) = 0.5:
= 4 + 6(1)(0.5) + 6(0.5) + 9
= 4 + 3 + 3 + 9
= 19
The correct calculation for |2u+3v| is the square root of 19, which is approximately 4.36 (not 4.71).
b) |4u-5v|^2 = (4u - 5v) · (4u - 5v)
= (4u) · (4u) + (4u) · (-5v) + (-5v) · (4u) + (-5v) · (-5v)
= 16(u · u) - 20(u · v) - 20(v · u) + 25(v · v)
= 16(|u|^2) - 20(u · v) - 20(v · u) + 25(|v|^2)
= 16(1) - 20(u · v) - 20(v · u) + 25(1)
= 16 - 20(1)(cos(60)) - 20(cos(60)) + 25
Again using cos(60) = 0.5:
= 16 - 20(1)(0.5) - 20(0.5) + 25
= 16 - 10 - 10 + 25
= 21
The correct calculation for |4u-5v| is the square root of 21, which is approximately 4.58 (not 3.22).
So, the error in your calculations is squaring only the magnitudes of the coefficients instead of squaring the entire vector.