VQ - Scalar Multiples of Vectors Lesson

Scalar Multiples of Vectors

We often want to multiply a vector by a scalar, or number. This just means we are changing the size of the vector!

Vector a
Let's say we multiply vector a by a scale factor of 2, we can see that 2a is twice as long as a, but still headed in the same direction!
What if multiply vector a by 1/2 then 1/2 a is half as long as vector a, but still headed in the same direction!
The only time a scalar multiple can change the direction of a vector is if that number is negative. Take a look at -2a

So, you can see that to calculate the magnitude of a vector multiplied by a scalar is to multiply that number by the magnitude of the vector. To summarize, the magnitude of a vector, v, multiplied by a scalar, c, gives us:

equation image indicator . $LaTeX: \lVert{cv}\rVert=\left|c\right|\cdot \lVert{v}\rVert$ $\lVert{cv}\rVert=\left|c\right|\cdot \lVert{v}\rVert$ .
In other words, the absolute value of c is multiplied by the magnitude of vector v.

Now it's time for you try some problems.

Given the components of the vectors below, find the requested scalar multiples.

Problem: Given $LaTeX: \vec{a}=<-2, 3>$ $\vec{a}=<-2, 3>$ , find 3a.

Solution: $LaTeX: 3\vec{a}=<-6, 9>$ $3\vec{a}=<-6, 9>$

Problem: Given $LaTeX: \vec{b}=<1, 4>$ $\vec{b}=<1, 4>$ , find ½b

Solution: $LaTeX: \frac{1}{2}\vec{b}=<\frac{1}{2}, 2>$ $\frac{1}{2}\vec{b}=<\frac{1}{2}, 2>$

Problem: Given $LaTeX: \vec{d}=<0, 2>$ $\vec{d}=<0, 2>$ , find -4d

Solution: $LaTeX: -4\vec{d}=<0 -8>$ $-4\vec{d}=<0 -8>$

Problem: Given $LaTeX: \vec{a}=<5, -3>$ $\vec{a}=<5, -3>$ , find 2a

Solution: $LaTeX: -2\vec{a}=<10, -6>$ $-2\vec{a}=<10, -6>$

Vectors can be represented as vertical matrices. So, the vector equation image indicator $LaTeX: \left<2,\:7\right>$ $\left<2,\:7\right>$ would be written as the matrix $LaTeX: \begin{bmatrix} 2\\ 7 \end{bmatrix}$ $\begin{bmatrix} 2\\ 7 \end{bmatrix}$ .

So, what if we had a vector that had 3 components, equation image indicator $LaTeX: \left<2,\:1,\:3\right>$ $\left<2,\:1,\:3\right>$ meaning the vector moved 2 units along the x-axis, 1 unit along the y-axis, and 3 units along the z-axis, we would write it as the matrix $LaTeX: \begin{bmatrix} 2\\ 1 \\ 3 \end{bmatrix}$ $\begin{bmatrix} 2\\ 1 \\ 3 \end{bmatrix}$ .

Write the vectors below as matrices.

Problem: $<-2, 0, 4>$ text annotation indicator

Solution: $LaTeX: \begin{bmatrix} -2 \\ 0 \\ 4 \end{bmatrix}$ $\begin{bmatrix} -2 \\ 0 \\ 4 \end{bmatrix}$

Problem: LaTeX: <3, -7> $<3, -7>$ text annotation indicator

Solution: $LaTeX: \begin{bmatrix} 3 \\ -7 \end{bmatrix}$ $\begin{bmatrix} 3 \\ -7 \end{bmatrix}$

Problem: LaTeX: <-1, 5> $<-1, 5>$

Solution: $LaTeX: \begin{bmatrix} -1 \\ 5 \end{bmatrix}$ $\begin{bmatrix} -1 \\ 5 \end{bmatrix}$

Problem: LaTeX: <1, -3, 6> $<1, -3, 6>$ text annotation indicator

Solution: $LaTeX: \begin{bmatrix} 1 \\ -3 \\ 6 \end{bmatrix}$ $\begin{bmatrix} 1 \\ -3 \\ 6 \end{bmatrix}$

We can use matrices to transform vectors. We talked about this a little bit during our matrix unit, but let's take a closer look at it!

You'll explore transforming vectors more in the handout below, but for now – let's practice multiplying a few vectors.

Problem: Given the matrix $LaTeX: \begin{bmatrix} -1 & 1 \\ 0 & 2\\ \end{bmatrix}$ $\begin{bmatrix} -1 & 1 \\ 0 & 2\\ \end{bmatrix}$ multiply to the vector LaTeX: <3, 1> $<3, 1>$

Solution: $LaTeX: \begin{bmatrix} -2\\ 2 \end{bmatrix}$ $\begin{bmatrix} -2\\ 2 \end{bmatrix}$

Problem: Given the matrix $LaTeX: \begin{bmatrix} 1 & -1 \\ 3 & 0\\ \end{bmatrix}$ $\begin{bmatrix} 1 & -1 \\ 3 & 0\\ \end{bmatrix}$ multiply to the vector LaTeX: <4, -3> $<4, -3>$

Solution: $LaTeX: \begin{bmatrix} 7 \\ 12 \end{bmatrix}$ $\begin{bmatrix} 7 \\ 12 \end{bmatrix}$

Problem: Given the matrix $LaTeX: \begin{bmatrix} 4 & 4 \\ -1 & 2\\ \end{bmatrix}$ $\begin{bmatrix} 4 & 4 \\ -1 & 2\\ \end{bmatrix}$ multiply to the vector LaTeX: <2, 1> $<2, 1>$

Solution: $LaTeX: \begin{bmatrix} 12 \\ 0 \end{bmatrix}$ $\begin{bmatrix} 12 \\ 0 \end{bmatrix}$

IMAGES CREATED BY GAVS