Vector Decomposition

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:

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

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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 \( \vec{c}? \)

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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 \(x+y?\)

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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 \(P?\)

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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 \(\vec{x}-\vec{y}?\)

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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.