Find the unit vector of direction for the following vector quantities.
(a) force F = (9.0î − 6.0ĵ) N
F hat =
(b) displacement D = (−4.0î − 8.0ĵ) m
D hat =
(c) velocity
v = (−3.50î + 6.50ĵ) m/s
v hat =
Find the unit vector of direction for the following vector quantities.
(a) force F = (9.0î − 6.0ĵ) N
|F| = sqrt (81 + 36) = sqrt 117 = 10.817
so
(9/10.817)i - (6/10.817)j
which is
0.832 i - 0.555 j
==========
check .83^2 + .56^2 = .69 + .31 which is about one :)
Do the others the same way
(a) F hat = "Oh, the force of humor! The unit vector of F is simply F divided by its magnitude. So, let's calculate the magnitude of F first. F has a magnitude of √(9.0^2 + (-6.0)^2) N. Now we can divide each component of F by its magnitude to find the unit vector of direction. So, F hat = (√(9.0^2 + (-6.0)^2)/9.0)î + (√(9.0^2 + (-6.0)^2)/(-6.0))ĵ N, or simplifying it a bit, F hat = (√117/9)î + (√117/(-6))ĵ N."
(b) D hat = "Get ready for a funny unit vector! The same method applies here. Let's calculate the magnitude of D first. D has a magnitude of √((-4.0)^2 + (-8.0)^2) m. Now we divide each component of D by its magnitude to find the unit vector of direction. So, D hat = (√((-4.0)^2 + (-8.0)^2)/(-4.0))î + (√((-4.0)^2 + (-8.0)^2)/(-8.0))ĵ m. Simplifying, D hat = (√80/(-4))î + (√80/(-8))ĵ m. Oh, the humor of vectors!"
(c) v hat = "Ready for a speedily funny unit vector? Let's calculate the magnitude of v first. v has a magnitude of √((-3.50)^2 + 6.50^2) m/s. Now divide each component of v by its magnitude to find the unit vector of direction. So, v hat = (√((-3.50)^2 + 6.50^2)/(-3.50))î + (√((-3.50)^2 + 6.50^2)/6.50)ĵ m/s. Simplifying, we have v hat = (√55/(-3.50))î + (√55/6.50)ĵ m/s. Voila, the ha-ha of velocity!"
To find the unit vector of direction, also known as the direction vector or the hat vector, we simply divide the given vector by its magnitude.
(a) For the force vector F = (9.0î - 6.0ĵ) N,
Magnitude of F = √(9.0^2 + (-6.0)^2) = √(81 + 36) = √117 ≈ 10.82
Therefore, the unit vector of direction for F is:
F hat = (9.0î - 6.0ĵ) N / 10.82 ≈ (0.83î - 0.55ĵ) N
(b) For the displacement vector D = (-4.0î - 8.0ĵ) m,
Magnitude of D = √((-4.0)^2 + (-8.0)^2) = √(16 + 64) = √80 ≈ 8.94
Therefore, the unit vector of direction for D is:
D hat = (-4.0î - 8.0ĵ) m / 8.94 ≈ (-0.45î - 0.89ĵ) m
(c) For the velocity vector v = (-3.50î + 6.50ĵ) m/s,
Magnitude of v = √((-3.50)^2 + 6.50^2) = √(12.25 + 42.25) = √54.5 ≈ 7.38
Therefore, the unit vector of direction for v is:
v hat = (-3.50î + 6.50ĵ) m/s / 7.38 ≈ (-0.48î + 0.88ĵ) m/s
To find the unit vector of direction for each vector quantity, we need to divide each vector by its magnitude.
(a) To find the unit vector for force F = (9.0î − 6.0ĵ) N, we first need to find the magnitude of F:
|F| = √(9.0^2 + (-6.0)^2)
= √(81 + 36)
= √117
Now, we can calculate the unit vector F hat by dividing F by its magnitude:
F hat = F / |F|
= (9.0î − 6.0ĵ) N / √117
(b) To find the unit vector for displacement D = (−4.0î − 8.0ĵ) m, we first need to find the magnitude of D:
|D| = √((-4.0)^2 + (-8.0)^2)
= √(16 + 64)
= √80
Now, we can calculate the unit vector D hat by dividing D by its magnitude:
D hat = D / |D|
= (−4.0î − 8.0ĵ) m / √80
(c) To find the unit vector for velocity v = (−3.50î + 6.50ĵ) m/s, we first need to find the magnitude of v:
|v| = √((-3.50)^2 + 6.50^2)
= √(12.25 + 42.25)
= √54.5
Now, we can calculate the unit vector v hat by dividing v by its magnitude:
v hat = v / |v|
= (−3.50î + 6.50ĵ) m/s / √54.5
Note: The unit vector represents the direction of the vector quantity and has a magnitude of 1. It is commonly denoted by adding a hat (^) symbol on top of the vector quantity, as shown above.