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Mathematical Expression Editor
Linear Combinations of Vectors
When studying vectors, the two main operations we have learned about are vector
addition and scalar multiplication. Both are involved in the important concept of a
linear combination of vectors.
A vector \(\vec {v}\) is said to be a linear combination of vectors \(\vec {v}_1, \vec {v}_2,\ldots , \vec {v}_n\) if
For example, \(\begin{bmatrix} -4\\9\\-10\\-1\end{bmatrix}\) is a linear combination of \(\begin{bmatrix} -1\\3\\-3\\0\end{bmatrix}\), \(\begin{bmatrix} 2\\0\\1\\4\end{bmatrix}\) and \(\begin{bmatrix} 0\\1\\-1\\1\end{bmatrix}\) because
In this section we will focus on vectors in \(\RR ^2\) and \(\RR ^3\).
Visualizing Linear Combinations in \(\RR ^2\) and \(\RR ^3\)
Let’s start by visualizing linear combinations of two vectors in \(\RR ^2\).
Answer the questions below using the GeoGebra interactive. To use the interactive,
you can
(a)
Change vectors \(\vec {v}\) and \(\vec {w}\) by dragging the tips of these vectors.
(b)
Change the coefficients \(k_1\) and \(k_2\) of the linear combination by using sliders.
(a)
Let \(\vec {w}=\begin{bmatrix}1\\2\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}1\\-1\end{bmatrix}\). Find \(k_1\) and \(k_2\) such that \(k_1\vec {w}+k_2\vec {v}=\begin{bmatrix}4\\-1\end{bmatrix}\)
\[k_1=\answer {1},\quad k_2=\answer {3}\]
(b)
Let \(\vec {w}=\begin{bmatrix}1\\2\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}-2\\0\end{bmatrix}\). Find \(k_1\) and \(k_2\) such that \(k_1\vec {w}+k_2\vec {v}=\begin{bmatrix}3\\-2\end{bmatrix}\)
\[k_1=\answer {-1},\quad k_2=\answer {-2}\]
(c)
Use the same vectors \(\vec {w}\) and \(\vec {v}\) as in the previous part. Do you think it is possible
to express any vector in \(\RR ^2\) as a linear combination of \(\vec {w}\) and \(\vec {v}\)? Yes, No
(d)
Let \(\vec {w}=\begin{bmatrix}4\\2\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}-2\\-1\end{bmatrix}\). Do you think it is possible to express any vector in \(\RR ^2\) as a linear
combination of \(\vec {w}\) and \(\vec {v}\)? Yes, No
Visualizing linear combinations of vectors in \(\RR ^3\) is more difficult than doing so in \(\RR ^2\). The
following Exploration will help you do this.
We will start by visualizing linear combinations of two vectors, \(\vec {u}\) and \(\vec {v}\), in \(\RR ^3\). To use
the interactive below, define vectors \(\vec {u}\) and \(\vec {v}\). Use sliders to change the coefficients \(k_1\)
and \(k_2\) of the linear combination. You will see the linear combination \(k_1\vec {u}+k_2\vec {v}\) as the
pink vector along the diagonal of the parallelogram determined by \(\vec {u}\) and
\(\vec {v}\).
RIGHT-CLICK and DRAG the left panel to rotate the image.
We will now consider three vectors. Define vectors \(\vec {u}\), \(\vec {v}\) and \(\vec {w}\). Use sliders to change the
coefficients \(k_1, k_2\) and \(k_3\) of the linear combination. The linear combination \(k_1\vec {u}+k_2\vec {v}+k_3\vec {w}\) is shown as the
pink vector along the diagonal of the parallelepiped.
RIGHT-CLICK and DRAG the left panel to rotate the image.
Geometry of Linear Combinations
Use geometry to express \(\begin{bmatrix}3\\-4\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\begin{bmatrix}-4\\2\end{bmatrix}\).
We are looking for \(\begin{bmatrix}3\\-4\end{bmatrix}\) to be the diagonal of a parallelogram determined by scalar
multiples of \(\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\begin{bmatrix}-4\\2\end{bmatrix}\).
Because a scalar multiple of a vector can point in the same direction as the vector or
in the opposite direction, we will start by drawing straight lines determined by the
two vectors.
The two lines that we drew will contain the sides of the parallelogram we are looking
for. To find the other two sides we will draw lines parallel to \(\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\begin{bmatrix}-4\\2\end{bmatrix}\) through the head
of vector \(\begin{bmatrix}3\\-4\end{bmatrix}\).
Now the parallelogram is clearly visible.
The last remaining task is to identify the sides of the parallelogram as scalar
multiples of \(\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\begin{bmatrix}-4\\2\end{bmatrix}\). Observe that vectors \(\begin{bmatrix}1\\-3\end{bmatrix}\) and \(\begin{bmatrix}2\\-1\end{bmatrix}\) determine the parallelogram.
Vector \(\begin{bmatrix}2\\-1\end{bmatrix}\) is half the length of \(\begin{bmatrix}-4\\2\end{bmatrix}\) and points in the opposite direction, while the vector \(\begin{bmatrix}1\\-3\end{bmatrix}\) is
the same length as \(\begin{bmatrix}-1\\3\end{bmatrix}\) and also points in the opposite direction.
We write \(\begin{bmatrix}3\\-4\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\begin{bmatrix}-4\\2\end{bmatrix}\) as follows
The method we used in Example 4 to express the given vector as a linear
combination of two other vectors is sufficiently useful that we summarize the
steps.
Given two non-collinear vectors \(\vec {u}\) and \(\vec {v}\) in \(\RR ^2\), and a vector \(\vec {w}\), we can express \(\vec {w}\) as a linear
combination of \(\vec {u}\) and \(\vec {v}\) as follows:
(a)
Draw lines \(L_{\vec {u}}\) and \(L_{\vec {v}}\) determined by \(\vec {u}\) and \(\vec {v}\), respectively.
(b)
Through the head of vector \(\vec {w}\), draw lines \(P_{\vec {u}}\) and \(P_{\vec {v}}\), parallel to \(L_{\vec {u}}\) and \(L_{\vec {v}}\), respectively.
(c)
Let \(A\) be the point of intersection of \(P_{\vec {u}}\) and \(L_{\vec {v}}\).
(d)
Let \(B\) be the point of intersection of \(P_{\vec {v}}\) and \(L_{\vec {u}}\).
(e)
Let \(O\) denote the origin. Then \(\overrightarrow {OA}=k_1\vec {v}\) and \(\overrightarrow {OB}=k_2\vec {u}\) for some scalars \(k_1\) and \(k_2\).
(f)
We have \(\vec {w}=k_2\vec {u}+k_1\vec {v}\).
This GeoGebra interactive will allow you to go through the steps given in
Procedure 6 for a combination of vectors of your choice. To use the interactive
(a)
Enter components of vectors \(\vec {u}\) and \(\vec {v}\).
(b)
Enter components of vector \(\vec {w}\) that you want to express as a linear
combination of \(\vec {u}\) and \(\vec {v}\).
(c)
Use the navigation bar to go through the steps of Procedure 6
From Geometry to Algebra of Linear Combinations
One of the stipulations in Procedure 6 is that vectors \(\vec {u}\) and \(\vec {v}\) should be non-collinear.
You can use the interactive in Exploration to investigate what happens when \(\vec {u}\) and \(\vec {v}\)
are collinear. The following example examines what happens from a geometric as well
as an algebraic standpoint.
Can the vector \(\begin{bmatrix}3\\2\end{bmatrix}\) be written as a linear combination of vectors \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\)?
We will start
with a geometric approach.
Observe that \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\) are scalar multiples of each other and lie on the same
line.
A linear combination of \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\) has the form:
This shows that all linear combinations of \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\) will be scalar multiples of \(\begin{bmatrix}-3\\1\end{bmatrix}\), and
therefore lie on the same line as \(\begin{bmatrix}-3\\1\end{bmatrix}\). Since \(\begin{bmatrix}3\\2\end{bmatrix}\) does not lie on the line determined by \(\begin{bmatrix}-3\\1\end{bmatrix}\) it
cannot be expressed as a linear combination of \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\).
We can also address this question algebraically. To express \(\begin{bmatrix}3\\2\end{bmatrix}\) as a linear combination of
\(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\), we need to solve the equation.
When you try to solve this system, you will find that the system is inconsistent. Thus,
\(\begin{bmatrix}3\\2\end{bmatrix}\) cannot be written as a linear combination of \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\).
We know that there is no way to express \(\begin{bmatrix}3\\2\end{bmatrix}\) as a linear combination of vectors \(\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\begin{bmatrix}6\\-2\end{bmatrix}\).
What would happen if we tried to apply Procedure 6 to these vectors? You can use
the GeoGebra interactive in Exploration to find out.
Express \(\begin{bmatrix}2\\4\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}2\\1\end{bmatrix}\) and \(\begin{bmatrix}2\\-2\end{bmatrix}\). Interpret your results geometrically.
This amounts to solving a system of linear equations
\begin{align*} 2a+2b&=2\\ a-2b&=4 \end{align*}
Use your favorite method to solve this system. (Hint: adding the second equation to
the first will work well for this system.) You will find that \(a=2\) and \(b=-1\). Now we can write \(\begin{bmatrix}2\\4\end{bmatrix}\)
as a linear combination of \(\begin{bmatrix}2\\1\end{bmatrix}\) and \(\begin{bmatrix}2\\-2\end{bmatrix}\) as follows:
Geometrically speaking, this means that the vector \(\begin{bmatrix}2\\4\end{bmatrix}\) is the diagonal of the
parallelogram determined by \(2\begin{bmatrix}2\\1\end{bmatrix}\) and \((-1)\begin{bmatrix}2\\-2\end{bmatrix}\). The original vectors \(\begin{bmatrix}2\\1\end{bmatrix}\) and \(\begin{bmatrix}2\\-2\end{bmatrix}\) are shown below
together with the parallelogram and its diagonal.
If possible, express \(\begin{bmatrix}7\\4\\-5\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}1\\-2\\1\end{bmatrix}\) and \(\begin{bmatrix}3\\0\\-1\end{bmatrix}\).
We are looking for coefficients \(a\)
and \(b\) such that
We conclude that \(\begin{bmatrix}7\\4\\-5\end{bmatrix}\) is a linear combination of \(\begin{bmatrix}1\\-2\\1\end{bmatrix}\) and \(\begin{bmatrix}3\\0\\-1\end{bmatrix}\), and write:
Set up a system of equations that can be used to express \(\begin{bmatrix}2\\-1\\3\\0\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}1\\0\\4\\-2\end{bmatrix}\),
\(\begin{bmatrix}-2\\-1\\1\\-1\end{bmatrix}\), \(\begin{bmatrix}0\\4\\-3\\1\end{bmatrix}\) and \(\begin{bmatrix}1\\1\\-1\\4\end{bmatrix}\), or to determine that such a combination does not exist. Do not solve the
system.
We are looking for \(x_1\), \(x_2\), \(x_3\) and \(x_4\) such that
Solve a system of linear equations to express \(\begin{bmatrix}-1\\7\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}1\\2\end{bmatrix}\) and
\(\begin{bmatrix}-1\\1\end{bmatrix}\).
Use Procedure 6 to express \(\begin{bmatrix}-3\\0\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}2\\4\end{bmatrix}\) and \(\begin{bmatrix}-1\\1\end{bmatrix}\).
Use two different approaches (algebraic and geometric) to explain why
the vector \(\begin{bmatrix}5\\1\end{bmatrix}\) cannot be expressed as a linear combination of vectors \(\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\begin{bmatrix}-4\\2\end{bmatrix}\).
We have seen Procedure 6 applied to vectors in \(\RR ^2\). The same process can, in certain
cases be applied to vectors in \(\RR ^3\). Follow the steps in Procedure 6 to express one vector
as a linear combination of two given vectors.
The following GeoGebra interactive shows vectors \({\bf u}=\begin{bmatrix}3\\0\\-1\end{bmatrix}\), \({\bf v}=\begin{bmatrix}1\\-2\\1\end{bmatrix}\) and \({\bf w}=\begin{bmatrix}7\\4\\-5\end{bmatrix}\). Use the navigation bar to
click through the construction steps. RIGHT-CLICK and DRAG to rotate the
image.
(a)
Can \(\vec {w}\) be expressed as a linear combination of \(\vec {u}\) and \(\vec {v}\)?
No, because \(\vec {w}\) is not
between \(\vec {u}\) and \(\vec {v}\).Yes, because all three vectors are in the same plane. Yes,
because all three vectors are in the same plane, AND \(\vec {u}\) and \(\vec {v}\) are not collinear.
(b)
Use the navigation bar at the bottom of the interactive window to view
construction steps of Procedure 6 applied to vectors \(\vec {u}\), \(\vec {v}\) and \(\vec {w}\). (Right-click and
drag to rotate the image.) Use the final image to express \(\bf w\) as a linear
combination of \(\bf v\) (blue) and \(\bf u\) (red).
The following GeoGebra interactive shows vectors \({\bf v}=\begin{bmatrix}1\\-2\\1\end{bmatrix}\), \({\bf u}=\begin{bmatrix}3\\0\\-1\end{bmatrix}\), and \({\bf w}=\begin{bmatrix}7\\4\\0\end{bmatrix}\). RIGHT-CLICK and
DRAG to rotate the image. Use geometry to explain why \(\bf w\) cannot be expressed as a
linear combination of \(\bf v\) and \(\bf u\).
We can also show that \(\bf w\) is not a linear combination of \(\bf v\) and \(\bf u\) algebraically by
attempting to solve a system of equations corresponding to
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