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Unit Vectors - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
One problem you’ll probably see in your homework at some point is find the unit vector with the same direction as v which is equals to <-4, 3> or some vector like that. Unit vector, the definition of a unit vector is that a unit vector has a length of 1. Any vector that has length 1 is a unit vector, but what’s the length of v? It’s the square root of -4² plus 3², which is 16 plus 9 which is 25. Root 25 is 5. The length of v is 5.
In order to get a unit vector, I need a vector that points in the same direction as v, but has 1/5 of the length. My unit vector, I’ll call it u-hat is going to be 1/5 of v. 1/5 of <-4, 3>. That’s <-4/5, 3/5>. This suggests a general method for coming up with the unit vector. For any vector v, a unit vector u with the same direction is, u that is 1 over the magnitude of v times the vector v. So this is a scalar multiple that gives you a length of 1 but the same direction as vector v. Let’s try that out on some problems.
I want to find the unit vector with the same direction as all of these vectors starting with <3, 3>. What’s the length of this vector? It’s the square root of 3² plus –3². So 9 plus 9, root 18 which is 3 root 2. That’s the length. And then the unit vector is going to be 1 over 3 root 2 times this vector. You pull this scalar inside and you get 3 over 3 root 2, -3 over 3 root 2. If you simplify this, you’ll get root 2 over 2, negative root 2 over 2. That’s the unit vector with the same direction as <3, -3>.
What about this guy? V is <-1, -2>. We first find the length. The length of v is the square root of -1² or 1 plus -2², 4. So this is root 5. The unit vector, u-hat, will be 1 root 5, 1 over the length times this vector. That’s going to be <-1 over root 5 and then -2 over root 5>. Your teacher will probably want you to rationalize the denominator. You’ll get negative root 5 over 5 and -2 root 5 over 5.
Finally, let’s try this one. V is <-5, 12>, what’s the magnitude? What’s the length? It’s the square root of -5², 25, plus 12², 144. 25 and 144 are 169 and root 169 is 13. The unit vector is going to be u-hat equals 1 over 13 times <-5, 12>. I distribute the 1/13 over the two components and I get <-5/ 13, 12/13>. That’s it.
It’s really easy to find the unit vector in the direction of any given vector. You first have to take the length or magnitude of the vector, and then you multiply by a scalar, which is the reciprocal of that length, and you get your unit vector.
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