3.3: Arithmetic on Vectors in 3-Dimensional Space
Addition and Subtraction of Vectors
To add or subtract two vectors, add or subtract their corresponding components.
Add the vectors \(\overrightarrow{u}=\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 4,\left.2,1\right\rangle \right.,\)
Solution
To add the vectors \(\overrightarrow{u}=\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 4,\left.2,1\right\rangle \right.,\) add their corresponding components.
\[\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 2+4,\left.5+2,\ 4+1\right\rangle \right.=\ \left\langle 6,\left.7,\ 5\right\rangle \right. \nonumber \] So, \(\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 6,\left.7,\ 5\right\rangle \right.\)
Now, graph this sum. Start at the origin.
Since the \(x-\) component is 6, move 6 units in the \(x-\) direction.
Since the \(y-\) component is 7, move 7 units in the \(y-\) direction.
Since the \(z-\) component is 5, move 5 units upward.
Subtract the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 4,\left.2,\ 1\right\rangle \right.\)
Solution
To subtract the vectors \(\overrightarrow{u}\boldsymbol{=}\left\langle 2,\ 5,\ \left.4\right\rangle \right.\) and \(\overrightarrow{v}\boldsymbol{=}\left\langle 4,\left.2,\ 1\right\rangle \right.\) subtract their corresponding components.
\[\overrightarrow{u}-\ \overrightarrow{v}\boldsymbol{=}\left\langle 2-4,\left.5-2,\ 4-1\right\rangle \right.\boldsymbol{=\ }\left\langle -2,\left.3,\ 3\right\rangle \right. \nonumber \]
So, \(\overrightarrow{u}-\ \overrightarrow{v}\boldsymbol{=}\left\langle -2,\left.3,\ 3\right\rangle \right.\)
Scalar Multiplication
Scalar multiplication is the multiplication of a vector by a real number (a scalar).
Suppose we let the letter \(k\) represent a real number and \(\overrightarrow{v}\) be the vector \(\left\langle x\right.,\left.y,z\right\rangle .\boldsymbol{\ }\) Then, the scalar multiple of the vector \(\overrightarrow{v}\) is
\[k \vec{v}=\langle k x, k y, k z\rangle \nonumber \]
Suppose \(\overrightarrow{u}=\left\langle -3,\left.-8,\ 5\right\rangle \right.\) and \(k=3\) .
Solution
Then \(k\overrightarrow{u}=\) 3 \(\overrightarrow{u}=3\left\langle -3,\left.-8,\ 5\right\rangle \right.=\ \left\langle 3(-3),\left.3\left(-8\right),3\eqref{GrindEQ__5_}\right\rangle \right.=\left\langle -9,\left.-24,\ 15\right\rangle \right.\)
Suppose \(\overrightarrow{v}=\left\langle 6,\left.3,-12\right\rangle \right.\) and \(k=\frac{-1}{3}\) .
Solution
Then \(k\overrightarrow{u}=\) \(\frac{-1}{3}\overrightarrow{u}=\frac{-1}{3}\left\langle 6,\left.3,\ -12\right\rangle \right.=\ \left\langle \frac{-1}{3}\left(6\right),\left.\frac{-1}{3}\left(3\right),\frac{-1}{3}(-12)\right\rangle \right.=\left\langle -2,\left.-1,\ 4\right\rangle \right.\)
Suppose \(\overrightarrow{u}=\left[ \begin{array}{c} \begin{array}{c} -2 \\ 6 \end{array} \\ 0 \end{array} \right]\) , \(\overrightarrow{v}=\left[ \begin{array}{c} 1 \\ 2 \\ -8 \end{array} \right]\) , and \(\overrightarrow{w}=\left[ \begin{array}{c} -3 \\ -1 \\ 2 \end{array} \right]\) . Find \(3\overrightarrow{u}+4\overrightarrow{v}-2\overrightarrow{w}.\)
Solution
Then \(3\overrightarrow{u}+4\overrightarrow{v}-2\overrightarrow{w}=3\left[ \begin{array}{c} \begin{array}{c} -2 \\ 6 \end{array} \\ 0 \end{array} \right]+4\left[ \begin{array}{c} 1 \\ 2 \\ -8 \end{array} \right]-2\left[ \begin{array}{c} -3 \\ -1 \\ 2 \end{array} \right]=\left[ \begin{array}{c} \begin{array}{c} -6 \\ 18 \end{array} \\ 0 \end{array} \right]+\left[ \begin{array}{c} 4 \\ 8 \\ -32 \end{array} \right]+\left[ \begin{array}{c} 6 \\ 2 \\ -4 \end{array} \right]=\left[ \begin{array}{c} 4 \\ 28 \\ -36 \end{array} \right] \nonumber \)
Using Technology
We can use technology to determine the value of adding or subtracting vectors.
Go to www.wolframalpha.com.
Suppose \(\overrightarrow{u}=\left[ \begin{array}{c} \begin{array}{c} -2 \\ 6 \end{array} \\ 0 \end{array} \right]\) , \(\overrightarrow{v}=\left[ \begin{array}{c} 1 \\ 2 \\ -8 \end{array} \right]\) , and \(\overrightarrow{w}=\left[ \begin{array}{c} -3 \\ -1 \\ 2 \end{array} \right]\) . Use WolframAlpha to find \(3\overrightarrow{u}+4\overrightarrow{v}-2\overrightarrow{w}.\) In the entry field enter evaluate \(3 *[-2,6,0]+4 *[1,2,-8]-2 *[-3,-1,2]\).
plus 4 times <1,2, negative 8> minus 2 times
WolframAlpha answers \((4,\ 28,\ -36)\) which is WolframAlpha’s notation for \(\left[ \begin{array}{c} 4 \\ 28 \\ -36 \end{array} \right]\) .
Try These
Add the vectors \(\vec{u}=\langle-3,4,6\rangle\) and \(\vec{v}=\langle 8,7,-5\rangle\).
- Answer
-
\(\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 5,\left.11,\ 1\right\rangle \right.\)
Subtract the vector \(\vec{v}=\langle 8,7,-5\rangle\) from the vector \(\vec{u}=\langle-3,4,6\rangle\).
- Answer
-
\(\overrightarrow{u}-\ \overrightarrow{v}=\left\langle -11,\left.-3,\ 11\right\rangle \right.\)
Given the three vectors, \(\vec{u}=\langle 2,4,-5\rangle, \vec{v}=\langle-3,4,-8\rangle\), and \(\vec{w}=\langle 0,1,2\rangle\), find \(2 \vec{u}+3 \vec{v}-4 \vec{w}\).
- Answer
-
\(2\overrightarrow{u}+3\overrightarrow{v}-4\overrightarrow{w}\ =\left\langle -5,\left.16,\ -42\right\rangle \right.\)
Suppose \(\vec{u}=\left[\begin{array}{c}
3 \\
4 \\
-2
\end{array}\right], \vec{v}=\left[\begin{array}{c}
-1 \\
6 \\
4
\end{array}\right]\), and \(\vec{w}=\left[\begin{array}{l}
0 \\
5 \\
2
\end{array}\right]\), find \(4 \vec{u}-4 \vec{v}-\vec{w}\)
- Answer
-
\(4\overrightarrow{u}-4\overrightarrow{v}-\overrightarrow{w}=\left\langle 16,\left.-13,\ -26\right\rangle \right.\)