3.3: Arithmetic on Vectors in 3-Dimensional Space

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Addition and Subtraction of Vectors

To add or subtract two vectors, add or subtract their corresponding components.

Example $\PageIndex{1}$

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, $\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 6,\left.7,\ 5\right\rangle \right.$

$A diagram showing two vectors in a 3-dimensional coordinate system. In the xyz plane, we have two vectors. We have vector u that has a starting point of (0,0,0) and a terminal point of (2,5,4). We have vector v that has a starting point of (0,0,0) and a terminal point of (4,2,1). There are lines that connect these vectors to the x-axis, the y-axis, and the z-axis.$

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.

$A diagram showing two vectors in a 3-dimensional coordinate system and the graph of their sum. In the xyz plane, we have three vectors. We have vector u that has a starting point of (0,0,0) and a terminal point of (2,5,4). We have vector v that has a starting point of (0,0,0) and a terminal point of (4,2,1). Then, we have a vector u + v has a starting point of (0,0,0) and the terminal point is (2,5,1). There are lines that connect these vectors to the x-axis, the y-axis, and the z-axis.$

Example $\PageIndex{2}$

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

Definition: Term

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

\[k \vec{v}=\langle k x, k y, k z\rangle \nonumber \]

Example $\PageIndex{3}$

Suppose $\overrightarrow{u}=\left\langle -3,\left.-8,\ 5\right\rangle \right.$ and $k=3$.

Solution

Then $k\overrightarrow{u}=$ 3$\overrightarrow{u}=3\left\langle -3,\left.-8,\ 5\right\rangle \right.=\ \left\langle 3(-3),\left.3\left(-8\right),3\eqref{GrindEQ__5_}\right\rangle \right.=\left\langle -9,\left.-24,\ 15\right\rangle \right.$

Example $\PageIndex{4}$

Suppose $\overrightarrow{v}=\left\langle 6,\left.3,-12\right\rangle \right.$ and $k=\frac{-1}{3}$.

Solution

Then $k\overrightarrow{u}=$ $\frac{-1}{3}\overrightarrow{u}=\frac{-1}{3}\left\langle 6,\left.3,\ -12\right\rangle \right.=\ \left\langle \frac{-1}{3}\left(6\right),\left.\frac{-1}{3}\left(3\right),\frac{-1}{3}(-12)\right\rangle \right.=\left\langle -2,\left.-1,\ 4\right\rangle \right.$

Example $\PageIndex{5}$

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 $3 *[-2,6,0]+4 *[1,2,-8]-2 *[-3,-1,2]$.

This screenshot from WolframAlpha shows an equation with scalar multiplication and vector addition and subtraction. We have 3 times <negative 2,6,0 plus 4 times <1,2, negative 8> minus 2 times . The result is <4,28, negative 36>." src="/@api/deki/files/97440/clipboard_e674088772cf81d9ce229ad6440ed593c.png">

WolframAlpha answers $(4,\ 28,\ -36)$ which is WolframAlpha’s notation for $\left[ \begin{array}{c} 4 \\ 28 \\ -36 \end{array} \right]$.

Try These

Exercise $\PageIndex{1}$

Add the vectors $\vec{u}=\langle-3,4,6\rangle$ and $\vec{v}=\langle 8,7,-5\rangle$.

Answer: $\overrightarrow{u}\ +\ \overrightarrow{v}=\left\langle 5,\left.11,\ 1\right\rangle \right.$

Exercise $\PageIndex{2}$

Subtract the vector $\vec{v}=\langle 8,7,-5\rangle$ from the vector $\vec{u}=\langle-3,4,6\rangle$.

Answer: $\overrightarrow{u}-\ \overrightarrow{v}=\left\langle -11,\left.-3,\ 11\right\rangle \right.$

Exercise $\PageIndex{3}$

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: $2\overrightarrow{u}+3\overrightarrow{v}-4\overrightarrow{w}\ =\left\langle -5,\left.16,\ -42\right\rangle \right.$

Exercise $\PageIndex{4}$

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: $4\overrightarrow{u}-4\overrightarrow{v}-\overrightarrow{w}=\left\langle 16,\left.-13,\ -26\right\rangle \right.$

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