Every vector can be thought of as the sum of two or more other vectors.
Consider the three vectors in the figure above. Since A C → \overrightarrow{AC} A C A B → \overrightarrow{AB} A B B C → , \overrightarrow{BC} , BC ,
A C → = A B → + B C → . \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} . A C = A B + 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 ) , O=(0,0,0), A=(1,0,0), B=(0,1,0), C=(0,0,1), O = ( 0 , 0 , 0 ) , A = ( 1 , 0 , 0 ) , B = ( 0 , 1 , 0 ) , C = ( 0 , 0 , 1 ) , D = ( 3 , 4 , 5 ) , D=(3,4,5) , D = ( 3 , 4 , 5 ) , a + b + c , a+b+c, a + b + c , a , b , a, b, a , b , c c c
O D → = a ⋅ O A → + b ⋅ O B → + c ⋅ O C → ? \overrightarrow{OD} = a\cdot\overrightarrow{OA}+b\cdot\overrightarrow{OB}+c\cdot\overrightarrow{OC}? O D = a ⋅ O A + b ⋅ OB + c ⋅ OC ?
We have
O D → = ( 3 , 4 , 5 ) , O A → = ( 1 , 0 , 0 ) , O B → = ( 0 , 1 , 0 ) , O C → = ( 0 , 0 , 1 ) . \begin{array}{c}&\overrightarrow {OD} = (3,4,5), &\overrightarrow {OA} = (1,0,0), &\overrightarrow {OB} = (0,1,0), &\overrightarrow {OC} = (0,0,1).\end{array} O D = ( 3 , 4 , 5 ) , O A = ( 1 , 0 , 0 ) , OB = ( 0 , 1 , 0 ) , OC = ( 0 , 0 , 1 ) .
Then,
O D → = a ⋅ O A → + b ⋅ O B → + c ⋅ O C → ( 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 ) ⇒ a = 3 , b = 4 , c = 5. \begin{aligned}
\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{aligned} O D ( 3 , 4 , 5 ) ⇒ a = a ⋅ O A + b ⋅ OB + c ⋅ OC = a ( 1 , 0 , 0 ) + b ( 0 , 1 , 0 ) + c ( 0 , 0 , 1 ) = ( a , 0 , 0 ) + ( 0 , b , 0 ) + ( 0 , 0 , c ) = ( a , b , c ) = 3 , b = 4 , c = 5.
Hence, a + b + c = 3 + 4 + 5 = 12. a+b+c = 3+4+5 = 12. a + b + c = 3 + 4 + 5 = 12. □ _\square □
Observe that O D → \overrightarrow{OD} O D O A → , O B → , \overrightarrow{OA},\overrightarrow{OB}, O A , OB , O C → \overrightarrow{OC} OC
If a ⃗ = ( 1 , 1 , 0 ) , b ⃗ = ( 1 , 0 , 1 ) \vec{a} = (1,1,0), \vec{b} = (1,0,1) a = ( 1 , 1 , 0 ) , b = ( 1 , 0 , 1 ) c ⃗ = ( 0 , 1 , 1 ) , \vec{c} = (0,1,1), c = ( 0 , 1 , 1 ) , d ⃗ = ( 5 , 6 , 7 ) \vec{d} = (5,6,7) d = ( 5 , 6 , 7 ) a ⃗ , b ⃗ \vec{a}, \vec{b} a , b c ⃗ ? \vec{c}? c ?
Let d ⃗ = x a ⃗ + y b ⃗ + z c ⃗ . \vec{d}= x \vec{a} + y \vec{b} + z \vec{c} . d = x a + y b + z c .
d ⃗ = x a ⃗ + y b ⃗ + z c ⃗ ( 5 , 6 , 7 ) = x ( 1 , 1 , 0 ) + y ( 1 , 0 , 1 ) + z ( 0 , 1 , 1 ) = ( x + y , x + z , y + z ) , \begin{aligned}
\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{aligned} d ( 5 , 6 , 7 ) = x a + y b + z c = x ( 1 , 1 , 0 ) + y ( 1 , 0 , 1 ) + z ( 0 , 1 , 1 ) = ( x + y , x + z , y + z ) ,
which implies
x + y = 5 x + z = 6 y + z = 7 ⇒ x = 2 , y = 3 , z = 4. \begin{aligned}
x+y &= 5 \\
x+z &= 6 \\
y+z &= 7 \\\\
\Rightarrow x=2, y&=3, z=4.
\end{aligned} x + y x + z y + z ⇒ x = 2 , y = 5 = 6 = 7 = 3 , z = 4.
Therefore,
d ⃗ = 2 a ⃗ + 3 b ⃗ + 4 c ⃗ . □ \vec{d} = 2 \vec{a} + 3 \vec{b} + 4 \vec{c}.\ _\square d = 2 a + 3 b + 4 c . □
If the vectors a ⃗ = ( 3 , − 2 , − 4 ) \vec{a} = (3,-2,-4) a = ( 3 , − 2 , − 4 ) b ⃗ = ( x + 1 , 8 , 2 y ) \vec{b} = (x+1,8,2y) b = ( x + 1 , 8 , 2 y ) x + y ? x+y? x + y ?
Since a ⃗ \vec{a} a b ⃗ \vec{b} b m m m b ⃗ = m a ⃗ \vec{b} = m \vec{a} b = m a
b ⃗ = m a ⃗ ( x + 1 , 8 , 2 y ) = m ( 3 , − 2 , − 4 ) = ( 3 m , − 2 m , − 4 m ) , \begin{aligned}
\vec{b} &= m\vec{a} \\
(x+1, 8, 2y) &= m(3,-2,-4) \\
&= (3m, -2m, -4m),
\end{aligned} b ( x + 1 , 8 , 2 y ) = m a = m ( 3 , − 2 , − 4 ) = ( 3 m , − 2 m , − 4 m ) ,
which implies
x + 1 = 3 m 8 = − 2 m 2 y = − 4 m ⇒ m = − 4 , x = − 13 , y = 8. \begin{aligned}
x+1 &= 3m\\
8&=-2m \\
2y&= -4m \\\\
\Rightarrow m&= -4, x=-13, y=8.
\end{aligned} x + 1 8 2 y ⇒ m = 3 m = − 2 m = − 4 m = − 4 , x = − 13 , y = 8.
Hence, x + y = − 13 + 8 = − 5. □ x+y = -13+8=-5. \ _ \square x + y = − 13 + 8 = − 5. □
The coordinates of points A , B , A,B, A , B , C C C ( 1 , 3 ) , ( 4 , 1 ) , (1,3), (4,1), ( 1 , 3 ) , ( 4 , 1 ) , ( 7 , 5 ) , (7,5) , ( 7 , 5 ) , P A → + P B → + P C → = 0 → , \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}=\overrightarrow{0}, P A + PB + PC = 0 , P ? P? P ?
Let the position vector of point P P P O P → = ( x , y ) . \overrightarrow{OP}=(x,y). OP = ( x , y ) .
P A → = ( 1 − x , 3 − y ) P B → = ( 4 − x , 1 − y ) P C → = ( 7 − x , 5 − y ) . \begin{aligned}
\overrightarrow{PA} &= (1-x,3-y) \\
\overrightarrow{PB} &= (4-x,1-y) \\
\overrightarrow{PC} &= (7-x, 5-y).
\end{aligned} P A PB PC = ( 1 − x , 3 − y ) = ( 4 − x , 1 − y ) = ( 7 − x , 5 − y ) .
Since we are given P A → + P B → + P C → = 0 → , \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{0} , P A + PB + PC = 0 ,
P A → + P B → + P C → = 0 → ( 1 − x , 3 − y ) + ( 4 − x , 1 − y ) + ( 7 − x , 5 − y ) = ( 0 , 0 ) ( 12 − 3 x , 9 − 3 y ) = ( 0 , 0 ) ⇒ x = 4 , y = 3. \begin{aligned}
\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{aligned} P A + PB + PC ( 1 − x , 3 − y ) + ( 4 − x , 1 − y ) + ( 7 − x , 5 − y ) ( 12 − 3 x , 9 − 3 y ) ⇒ x = 0 = ( 0 , 0 ) = ( 0 , 0 ) = 4 , y = 3.
Thus, the coordinates of point P P P P = ( 4 , 3 ) . P=(4,3). P = ( 4 , 3 ) . □ _\square □
If the vectors a ⃗ = ( 1 , − 2 ) \vec{a} = (1,-2) a = ( 1 , − 2 ) b ⃗ = ( 4 , − 5 ) \vec{b} = (4,-5) b = ( 4 , − 5 ) a ⃗ = x ⃗ + y ⃗ \vec{a}=\vec{x} + \vec{y} a = x + y b ⃗ = x ⃗ − 2 y ⃗ , \vec{b}= \vec{x} -2\vec{y}, b = x − 2 y , x ⃗ − y ⃗ ? \vec{x}-\vec{y}? x − y ?
We have
x ⃗ + y ⃗ = a ⃗ x ⃗ − 2 y ⃗ = b ⃗ ⇒ x ⃗ = 1 3 ( 2 a ⃗ + b ⃗ ) , y ⃗ = 1 3 ( a ⃗ − b ⃗ ) . \begin{aligned}
\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{aligned} x + y x − 2 y ⇒ x y = a = b = 3 1 ( 2 a + b ) , = 3 1 ( a − b ) .
Hence, x ⃗ − y ⃗ \vec{x} - \vec{y} x − y
x ⃗ − y ⃗ = 1 3 ( 2 a ⃗ + b ⃗ ) − 1 3 ( a ⃗ − b ⃗ ) = 1 3 ( a ⃗ + 2 b ⃗ ) = 1 3 a ⃗ + 2 3 b ⃗ = ( 1 3 , − 2 3 ) + ( 8 3 , − 10 3 ) = ( 3 , − 4 ) . □ \begin{aligned}
\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{aligned} x − y = 3 1 ( 2 a + b ) − 3 1 ( a − b ) = 3 1 ( a + 2 b ) = 3 1 a + 3 2 b = ( 3 1 , − 3 2 ) + ( 3 8 , − 3 10 ) = ( 3 , − 4 ) . □
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, O D → \overrightarrow{OD} O D O A → , O B → \overrightarrow{OA}, \overrightarrow{OB} O A , OB O C → . \overrightarrow{OC}. OC .
O D → = O A → + O B → + O C → . \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} . O D = O A + OB + OC .
One can define the unit vectors that point in each of the positive coordinate axes as follows:
x ^ = ( 1 , 0 , 0 ) y ^ = ( 0 , 1 , 0 ) z ^ = ( 0 , 0 , 1 ) . \begin{aligned} \hat{x} &= (1, 0, 0) \\
\hat{y} &= (0, 1, 0) \\
\hat{z} &= (0, 0, 1). \end{aligned} x ^ y ^ z ^ = ( 1 , 0 , 0 ) = ( 0 , 1 , 0 ) = ( 0 , 0 , 1 ) .
It follows that for any vector V ⃗ = ( v x , v y , v z ) \vec{V} = (v_x, v_y, v_z) V = ( v x , v y , v z )
V ⃗ = v x x ^ + v y y ^ + v z z ^ . \vec{V} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}. V = v x x ^ + v y y ^ + v z z ^ .
In this way, one frequently encounters vectors written in terms of the unit vectors x ^ \hat{x} x ^ y ^ \hat{y} y ^ z ^ \hat{z} z ^ i ^ = x ^ \hat{i} = \hat{x} i ^ = x ^ j ^ = y ^ \hat{j} = \hat{y} j ^ = y ^ k ^ = z ^ \hat{k} = \hat{z} k ^ = z ^
For instance, the vector ( − 1 , 2 , 3 ) (-1, 2, 3) ( − 1 , 2 , 3 ) − x ^ + 2 y ^ + 3 z ^ -\hat{x} + 2 \hat{y} + 3 \hat{z} − x ^ + 2 y ^ + 3 z ^
Note that the coordinate unit vectors are orthonormal . That is to say, x ^ ⋅ x ^ = y ^ ⋅ y ^ = z ^ ⋅ z ^ = 1 \hat{x} \cdot \hat{x} = \hat{y} \cdot \hat{y} = \hat{z} \cdot \hat{z} = 1 x ^ ⋅ x ^ = y ^ ⋅ y ^ = z ^ ⋅ z ^ = 1 x ^ ⋅ y ^ = y ^ ⋅ z ^ = x ^ ⋅ z ^ = 0 \hat{x} \cdot \hat{y} = \hat{y} \cdot \hat{z} = \hat{x} \cdot \hat{z} = 0 x ^ ⋅ y ^ = y ^ ⋅ z ^ = x ^ ⋅ z ^ = 0
When two vectors V ⃗ = ( v x , v y , v z ) \vec{V}= (v_x, v_y, v_z) V = ( v x , v y , v z ) W ⃗ = ( w x , w y , w z ) \vec{W} = (w_x, w_y, w_z) W = ( w x , w y , w z )
V ⃗ + W ⃗ = ( v x x ^ + v y y ^ + v z z ^ ) + ( w x x ^ + w y y ^ + w z z ^ ) = ( v x + w x ) x ^ + ( v y + w y ) y ^ + ( v z + w z ) z ^ , \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}, V + W = ( v x x ^ + v y y ^ + v z z ^ ) + ( w x x ^ + w y y ^ + w z z ^ ) = ( v x + w x ) x ^ + ( v y + w y ) y ^ + ( v z + w z ) z ^ ,
which is the same thing as ( v x + w x , v y + w y , v z + w z ) (v_x + w_x, v_y + w_y, v_z + w_z) ( v x + w x , v y + w y , v z + w z ) V ⃗ ⋅ W ⃗ \vec{V} \cdot \vec{W} V ⋅ W
V ⃗ ⋅ W ⃗ = ( v x x ^ + v y y ^ + v z z ^ ) ⋅ ( w x x ^ + w y y ^ + w z z ^ ) = v x w x x ^ ⋅ x ^ + v y w y y ^ ⋅ y ^ + v z w z z ^ ⋅ z ^ = v x w x + v y w y + v z w z , \begin{aligned}
\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{aligned} V ⋅ W = ( v x x ^ + v y y ^ + v z z ^ ) ⋅ ( w x x ^ + w y y ^ + w z z ^ ) = v x w x x ^ ⋅ x ^ + v y w y y ^ ⋅ y ^ + v z w z z ^ ⋅ z ^ = v x w x + v y w y + v z w z ,
as expected.
