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## Precalculus (Eureka Math/EngageNY)

### Course: Precalculus (Eureka Math/EngageNY)>Unit 2

Lesson 4: Topic D: Lesson 19: Directed line segments and vectors

# Scalar multiplication: magnitude and direction

Sal analyzes the magnitude and direction of vectors that are a result of scalar multiplication of a vector whose magnitude is given.

## Want to join the conversation?

• At Sal said that the magnitude can only be positive. But shouldn't that be non-negative?
• I'm sorry, I don't understand why that matters... what's the difference between positive and non-negative?
• what if the x and y have different multiples? Like (5x,8y)?
• No ones replied to you so I thought i'd give it a go...
Basically, you're not multiplying the vector by a scalar. You're multiplying its direction components by two separate scalars, if you want to think of it like that. It's the same principle however, you would just do √(5x)²+(8y)² ... we can't really imagine what this would look like however, as we don't know what the initial x,y components of the vector looks like. Although if you'd like an example:
Say vector 'V' has the direction components 3i + 4j then the magnitude of the vector would be √(3)²+(4)² which would equal 5. When applying your multiples (5 and 8) to each of the x, y components it would become √(15)²+(32)² which gives you 35.34 roughly...
So as you can see when multiplying the components by separate variables, it's no longer a simple scalar multiplication of the vector where we can just multiply the initial magnitude by say 3. Hope this clears things up.
• If a vector can have zero length, what would be its direction?
• Vector having zero length means it hasn't moved at all. Therefore, giving such vectors direction would be "pointless".
• how would you solve it if lets say that vector z was (-2x,3y)?
• First of all you would only have -2x and 3y if you were trying to write a Cartesian equation from the vector. In fact the vector z that you have specified is actually (-2,3) not (-2x,3y).
• For the second part of the first question, shouldn't the vector of z equal -10 not 10?
• No, because magnitude can't be negative.
• why does the vector have to be always with a letter and not a number?
• If we denoted vectors with numbers
the chance of mistaking the "vector number" for its magnitude will be pretty high, causing mistakes in the calculation
• magnitude of a vector is √x^2 + y^2
So, √x^2 + y^2 = 5
x^2 + y^2 = 25
3(x^2 + y^2) = 75
3x^2 +3y^2 = 75
√3x^2 +3y^2 = √75
ans = 5√3 = 8.66
What am i doing wrong here?
(1 vote)
• If you scale a vector by 3, <x, y> becomes <3x, 3y>. x is replaced by 3x, y is replaced by 3y.

So if the magnitude is √(x²+y²)=5, we replace x by 3x and y by 3y to get
√((3x)²+(3y)²)
=√(9x²+9y²)
=√(9(x²+y²))
=√9√(x²+y²)
=3√(x²+y²)
=3·5=15.
• At the very beginning, since the magnitude of vector v is equal to 5, could you say that (x,y) would be (5x, 5y)? Thanks.
(1 vote)
• No, because the magnitude of a vector is equal to the distance from the beginning point to the end point. The components x and y are the individual contributions in the x direction and y direction that would push the beginning point location to the end location. So the components x and y would be the sides of a right triangle and the magnitude of 5 is the length of the arrow (vector) that forms the hypotenuse. There are pretty much an infinite number of ways to get 5 as the magnitude. We can use trig to help us, but in your case, we can use x² + y² = v² from Pythagoras to figure out the components.
However, after we have said that the magnitude is 5 if the components are x and y, we cannot multiply the components by 5 and have the same vector. That would be a new, bigger vector.