# Vector Decomposition

## Summary

Every vector can be thought of as the sum of two or more other vectors.

Consider the three vectors in the figure above. Since \( \overrightarrow{AC} \) is composed of \( \overrightarrow{AB}\) and \( \overrightarrow{BC} ,\) the following expression is established:

\[ \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} .\]

**Vector decomposition** is the general process of breaking one vector into two or more vectors that add up to the original vector. The **component vectors** into which the original vector is decomposed are chosen based on specific details of the problem at hand.

Given the points \(O=(0,0,0), A=(1,0,0), B=(0,1,0), C=(0,0,1), \) and \(D=(3,4,5) ,\) what is the value of \(a+b+c,\) where \( a, b, \) and \( c \) satisfy the following equation:

\[ \overrightarrow{OD} = a\cdot\overrightarrow{OA}+b\cdot\overrightarrow{OB}+c\cdot\overrightarrow{OC}? \]

We have

\[ \begin{array} &\overrightarrow {OD} = (3,4,5), &\overrightarrow {OA} = (1,0,0), &\overrightarrow {OB} = (0,1,0), &\overrightarrow {OC} = (0,0,1).\end{array} \]

Then,

\[ \begin{align} \overrightarrow{OD} &= a\cdot\overrightarrow{OA}+b\cdot\overrightarrow{OB}+c\cdot\overrightarrow{OC} \\ (3,4,5) &= a(1,0,0) + b(0,1,0) + c(0,0,1) \\ &= (a,0,0) + (0,b,0) + (0,0,c) \\ &= (a,b,c) \\\\ \Rightarrow a &= 3, b = 4, c = 5. \end{align} \]

Hence, \(a+b+c = 3+4+5 = 12.\) \(_\square \)

Observe that \(\overrightarrow{OD}\) was decomposed into three vectors that are parallel to the three coordinate axes. Also notice that the vectors \(\overrightarrow{OA},\overrightarrow{OB},\) and \(\overrightarrow{OC}\) are the unit vectors of the coordinate space.

If \( \vec{a} = (1,1,0), \vec{b} = (1,0,1)\) and \(\vec{c} = (0,1,1),\) how can \( \vec{d} = (5,6,7) \) be expressed using \( \vec{a}, \vec{b}\) and \( \vec{c}? \)

Let \( \vec{d}= x \vec{a} + y \vec{b} + z \vec{c} .\) Then we have

\[ \begin{align} \vec{d} &= x \vec{a} + y \vec{b} + z\vec{c} \\ (5,6,7) &= x(1,1,0) + y(1,0,1) + z(0,1,1) \\ &= (x+y, x+z, y+z), \end{align} \]

which implies

\[ \begin{align} x+y &= 5 \\ x+z &= 6 \\ y+z &= 7 \\\\ \Rightarrow x=2, y&=3, z=4. \end{align} \]

Therefore,

\[ \vec{d} = 2 \vec{a} + 3 \vec{b} + 4 \vec{c}.\ _\square\]

If the vectors \( \vec{a} = (3,-2,-4) \) and \( \vec{b} = (x+1,8,2y) \) are parallel, what is the value of \(x+y?\)

Since \( \vec{a}\) and \( \vec{b} \) are parallel, a certain real number \(m\) which satisfies \( \vec{b} = m \vec{a} \) exists. Thus,

\[ \begin{align} \vec{b} &= m\vec{a} \\ (x+1, 8, 2y) &= m(3,-2,-4) \\ &= (3m, -2m, -4m), \end{align} \]

which implies

\[ \begin{align} x+1 &= 3m\\ 8&=-2m \\ 2y&= -4m \\\\ \Rightarrow m&= -4, x=-13, y=8. \end{align} \]

Hence, \( x+y = -13+8=-5. \ _ \square \)

The coordinates of points \(A,B,\) and \(C\) are \( (1,3), (4,1),\) and \((7,5) ,\) respectively. If \(\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}=\overrightarrow{0}, \) what are the coordinates of point \(P?\)

Let the position vector of point \(P\) be \(\overrightarrow{OP}=(x,y).\) Then we have

\[ \begin{align} \overrightarrow{PA} &= (1-x,3-y) \\ \overrightarrow{PB} &= (4-x,1-y) \\ \overrightarrow{PC} &= (7-x, 5-y). \end{align} \]

Since we are given \( \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{0} ,\) we have

\[ \begin{align} \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} &= \overrightarrow{0} \\ (1-x,3-y)+ (4-x,1-y)+(7-x, 5-y) &= (0,0) \\ (12-3x, 9-3y) &= (0,0) \\\\ \Rightarrow x&= 4, y= 3. \end{align} \]

Thus, the coordinates of point \(P\) are \( P=(4,3). \) \(_\square\)

If the vectors \( \vec{a} = (1,-2)\) and \( \vec{b} = (4,-5)\) satisfy \( \vec{a}=\vec{x} + \vec{y}\) and \(\vec{b}= \vec{x} -2\vec{y},\) what is \(\vec{x}-\vec{y}?\)

We have

\[ \begin{align} \vec{x} + \vec{y} &= \vec{a} \\ \vec{x} -2\vec{y} &= \vec{b} \\ \Rightarrow \vec{x} &= \frac{1}{3} \big(2 \vec{a} + \vec{b} \big), \\ \vec{y} &= \frac{1}{3} \big( \vec{a} - \vec{b}\big). \end{align} \]

Hence, \( \vec{x} - \vec{y} \) is

\[ \begin{align} \vec{x} - \vec{y} &= \frac{1}{3} \big(2 \vec{a} + \vec{b}\big) - \frac{1}{3} \big( \vec{a} - \vec{b}\big) \\ &= \frac{1}{3} \big( \vec{a}+2\vec{b} \big) \\ &= \frac{1}{3} \vec{a} + \frac{2}{3} \vec{b} \\ &= \left( \frac{1}{3}, -\frac{2}{3} \right) + \left( \frac{8}{3}, -\frac{10}{3}\right) \\ &= (3, -4).\ _\square \end{align} \]

One generally useful decomposition is with components that lie parallel to each of the coordinate axes.

The figure above shows the decomposition of a vector in a three-dimensional space. Observe that the red vector can be decomposed into the three green vectors that are parallel to the three coordinate axes. Thus, \( \overrightarrow{OD} \) is composed of the three vectors \( \overrightarrow{OA}, \overrightarrow{OB} \) and \( \overrightarrow{OC}.\) Therefore the following expression corresponds:

\[ \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} .\]

One can define the unit vectors that point in each of the positive coordinate axes as follows:

\[ \begin{align} \hat{x} &= (1, 0, 0) \\ \hat{y} &= (0, 1, 0) \\ \hat{z} &= (0, 0, 1). \end{align} \]

It follows that for any vector \( \vec{V} = (v_x, v_y, v_z) \), it holds that

\[ \vec{V} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}. \]

In this way, one frequently encounters vectors written in terms of the unit vectors \( \hat{x} \), \( \hat{y} \), and \( \hat{z} \). An alternate notation uses \( \hat{i} = \hat{x} \) , \( \hat{j} = \hat{y} \), and \( \hat{k} = \hat{z} \).

For instance, the vector \( (-1, 2, 3) \) can be expressed as \( -\hat{x} + 2 \hat{y} + 3 \hat{z} \).

Note that the coordinate unit vectors are orthonormal. That is to say, \( \hat{x} \cdot \hat{x} = \hat{y} \cdot \hat{y} = \hat{z} \cdot \hat{z} = 1 \) and \(\hat{x} \cdot \hat{y} = \hat{y} \cdot \hat{z} = \hat{x} \cdot \hat{z} = 0 \).

When two vectors \( \vec{V}= (v_x, v_y, v_z)\) and \( \vec{W} = (w_x, w_y, w_z) \) add,

\[ \vec{V}+\vec{W} = (v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) + (w_x \hat{x} + w_y \hat{y} + w_z \hat{z}) = (v_x + w_x) \hat{x} + (v_y + w_y) \hat{y} + (v_z + w_z) \hat{z}, \]

which is the same thing as \( (v_x + w_x, v_y + w_y, v_z + w_z) \). Similarly, when one takes the dot product \( \vec{V} \cdot \vec{W} \), one obtains

\[ \begin{align} \vec{V} \cdot \vec{W} &= (v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) \cdot (w_x \hat{x} + w_y \hat{y} + w_z \hat{z}) \\ &= v_x w_x \hat{x} \cdot \hat{x} + v_y w_y \hat{y} \cdot \hat{y} + v_z w_z \hat{z} \cdot \hat{z} \\ &= v_x w_x + v_y w_y + v_z w_z, \end{align} \]

as expected.

**Cite as:**Vector Decomposition.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/vector-decomposition/