5. Vectors

c. Scalar Multiplication

4. Linear Combinations

If we combine vector addition and scalar multiplication, we get a linear combination.

The linear combination of the vectors \(\vec u,\vec v,\cdots,\vec w\) with coefficients \(a,b,\cdots,c\) is the expression \[ a\vec u+b\vec v+\cdots+c\vec w \] Notice we can do this with any number of vectors and coefficients (as long as the numbers match).

For example, the linear combination of the vectors \(\vec p=\left\langle3,4\right\rangle\), \(\vec q=\left\langle-1,3\right\rangle\) and \(\vec r=\left\langle2,-2\right\rangle\) with coefficients \(2\), \(-3\), and \(4\) is \[\begin{aligned} 2\vec p-3\vec q+4\vec r &=2\left\langle3,4\right\rangle-3\left\langle-1,3\right\rangle+4\left\langle2,-2\right\rangle \\ &=\left\langle6,8\right\rangle+\left\langle3,-9\right\rangle+\left\langle8,-8\right\rangle =\left\langle17,-9\right\rangle \end{aligned}\]

Compute \(5\left\langle-2,1\right\rangle+3\left\langle1,2\right\rangle-2\left\langle2,4\right\rangle +\left\langle4,-3\right\rangle\).

\( 5\left\langle-2,1\right\rangle+3\left\langle1,2\right\rangle -2\left\langle2,4\right\rangle+\left\langle4,-3\right\rangle =\left\langle-7,0\right\rangle \)

\[\begin{aligned} 5&\left\langle-2,1\right\rangle+3\left\langle1,2\right\rangle -2\left\langle2,4\right\rangle+\left\langle4,-3\right\rangle \\ &=\left\langle-10,5\right\rangle+\left\langle3,6\right\rangle +\left\langle-4,-8\right\rangle+\left\langle4,-3\right\rangle \\ &=\left\langle-7,0\right\rangle \end{aligned}\]

Write the vector \(\vec v=\left\langle10,6\right\rangle\) as a linear combination of the vectors \(\vec p=\left\langle2,3\right\rangle\) and \(\vec q=\left\langle-1,3\right\rangle\). In other words, find numbers \(a\) and \(b\) satisfying: \[ \vec v=a\vec p+b\vec q \]

\( \vec v=4\vec p-2\vec q \)

We first write out the equation \(\vec v=a\vec p+b\vec q\): \[ \left\langle10,6\right\rangle =a\left\langle2,3\right\rangle+b\left\langle-1,3\right\rangle \] This is the two equations: \[\begin{aligned} 10&=2a-b \\ 6&=3a+3b \end{aligned}\] The second equation implies \(b=2-a\). We substitute this into the first equation which becomes: \[\begin{aligned} 10&=3a-2 \\ \end{aligned}\] Consequently, \(a=4\) and hence \(b=-2\). We conclude: \[ \vec v=4\vec p-2\vec q \]

We check: \[\begin{aligned} 4\vec p-2\vec q =4\left\langle2,3\right\rangle-2\left\langle-1,3\right\rangle =\left\langle10,6\right\rangle=\vec v \end{aligned}\]

You can use this Maplet to practice 2D vector algebra (requires Maple on the computer where this is executed):

2D Vector Algebra Rate It

5. Vectors

c. Scalar Multiplication

4. Linear Combinations

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