Video Transcript
Given that π is the vector negative five π’ minus six π£ and π is the vector negative four π’ minus six π£, where π’ and π£ are two perpendicular unit vectors, find the magnitude of two π minus two π.
In this question, weβre given two vectors, the vector π and the vector π, and weβre given these in terms of unit directional vectors π’ and π£. We need to use this to find the magnitude of the vector two π minus two π. Thereβs a few different ways we could approach this. However, because weβre given vector π and vector π in terms of unit directional vectors π’ and π£, the easiest way will be to find the vector two π minus two π in terms of these vectors. Then, weβll just be able to find the magnitude of this vector.
So, letβs start by calculating two π minus two π. Thatβs two times negative five π’ minus six π£ minus two times negative four π’ minus six π£. Now, we can see in this expression, weβre multiplying each of our two vectors π’ and π£ by a scalar. And to do this, we need to recall that to multiply a vector by a scalar, we just multiply each of our components by our scalar. So, in our first vector, weβre going to want to multiply each of our components by our scalar of two. So, in our first vector, we get the component of two multiplied by negative five, which is equal to negative 10. And in our second vector, we get a component of two times negative six, which is negative 12. So, we get negative 10π’ minus 12π£.
And in our second vector, we have two choices. We could multiply both of our components by two and then subtract the two vectors. Or we could multiply through by negative two and then add the two vectors. Weβll multiply through by negative two and then add our two vectors together. Remember, this means we need to multiply each of our components by negative two. This gives us negative two multiplied by negative four which is equal to positive eight and then negative two multiplied by negative six, which is equal to positive 12. So, we get eight π’ plus 12π£.
So, weβve shown two π minus two π is equal to negative 10π’ minus 12π£ plus eight π’ plus 12π£. And now, we want to simplify this expression. And we do this by adding the corresponding components together. We have negative 10π’ plus eight π’; this is going to be equal to negative two π’. And then, we have negative 12π£ plus 12π£; this is going to be equal to zero π£. And because our component is equal to zero, we donβt need to include this.
Now that we found an expression for two π minus two π, we can find its magnitude. Its magnitude will be equal to the magnitude of negative two π’. And thereβs a few different methods we could do this. For example, we could find the square root of the sum of the squares of the components of our vector. We could also do this graphically by choosing our coordinate system for our unit perpendicular vectors π’ and π£. However, the easiest way is probably to remember the following result. For any real constant π and vector π, the magnitude of ππ is equal to the modulus of π multiplied by the magnitude of π.
And all this is really telling us is if we scale a vector up by a factor of π, then weβre scaling its magnitude up by a factor of the modulus of π. In our case, our value of π is negative two and our vector π is just the unit directional vector π’. So, we can simplify this to get the modulus of negative two multiplied by the magnitude of π’. And in the question, weβre told π’ is a unit vector, so its magnitude is equal to one. And of course, the modulus of negative two is equal to two. So, it all simplifies to just give us two, which is our final answer.
Therefore, we were able to show if π is the vector negative five π’ minus six π£ and π is the vector negative four π’ minus six π£, where π’ and π£ are perpendicular unit vectors, then the magnitude of two π minus two π must be equal to two.
