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Unit Vectors - Problem 1 2,830 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

There are two ways we can express a vector algebraically. One is in component form, with the corner brackets, and a, b the horizontal and vertical components of the vector, and the other way is using the unit vectors i and j. It’s really easy to switch back and forth between these two forms, as long as you remember that i is the vector <1, 0> and j is vector <0, 1>. All you have to do is multiply through and you get <5, 0> minus <0, 9>. That’s <5, -9>. You can see that it’s a lot quicker than that.

Really these coefficients of i and j are the components, so you can go directly to this form, once you realize that. For example 8i plus 7 is <8, 1>. Those are the components. -16i is <-16, 0>.

This one’s a little trickier. We have a coefficient here. Notice that we have the magnitude of 3i minus 4j so that’s just going to be some real number and then 1 over that. Let me calculate that first.

The magnitude of 3i minus 4j, so square root of 3² plus -4². Those are the components, 3 and -4. We get 9 plus 16 which is 25, and root 25 is 5. This is really just 1/5 of 3i minus 4j. You can scalar multiply 3i minus 4j just the way you might expect. Just distribute this 1/5 over the two terms. You get 3/5i minus 4/5j. And then just convert to the component form, <3/5, -4/5> and that’s your final answer.

Again you’ve got your two forms, you’ve got component form with the corner brackets and the horizontal and vertical components. You’ve also got this i,j form with the unit vectors i and j. Remember the unit vectors are drawn with this little hat pronounced i-hat and j-hat.

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