Expressing a vector as the sum of two vectors

Question

I need to express v as the sum of two vectors, one parallel to w, the other perpendicular to w, where v is the vector <4, 3, 2> and w is the vector <1, 1, 1>.

This is what I have done:

Setup the equation that expresses v as a sum of two vectors where a is parallel to w and b is perpendicular to w.

v = a + b

Find initial value of a using the formula a = kw where k is a constant.

a = kw

= k * <1, 1, 1>

= (k, k, k)

Find initial value of b using the formula b = v - a.

b = v - a

= <4, 3, 2> - (k, k, k)

= <4 - k, 3 - k, 2 - k>

Solve dot product of b and w equal to 0 to find k and thus find the vectors a and b.

b · w = 0

<4 - k, 3 - k, 2 - k> · <1, 1, 1> = 0

4 - k + 3 - k + 2 - k = 0

9 - 3k = 0

-3k = -9

k = 3

So a = <3, 3, 3> and b = <1, 0, -1>

Therefore, v as the sum of two vectors is <3, 3, 3> + <1, 0, -1>.

Would this be the correct?

Answer 1

The answer you arrive at is correct, but you can arrive at it with less steps by decomposing $\textbf{v}$ into its horizontal and vertical components via projection: $$ \textbf{v} = {\rm proj}_{\textbf{w}}\textbf{v} + (\textbf{v} -{\rm proj}_{\textbf{w}}\textbf{v}) \quad \left(=\textbf{a}+\textbf{b}\right)$$ We arrive at the same answers you had: $$ {\rm proj}_{\textbf{w}}\textbf{v} = \frac{\textbf{v}\cdot \textbf{w}}{\textbf{w}\cdot \textbf{w}}\textbf{w} = \frac{9}{3}(1,1,1)=(3,3,3) = \textbf{a}$$ We see that $\textbf{a}$ is parallel to $\textbf{w}$ since it is a scalar multiple. $$ \textbf{v} - {\rm proj}_{\textbf{w}}\textbf{v} = (4,3,2)-(3,3,3) = (1,0,-1) = \textbf{b}$$ We see that $\textbf{b}$ is perpendicular to $\textbf{w}$ since $\textbf{b}\cdot\textbf{w}=0$, and as desired: $(3,3,3)+(1,0,-1) = (4,3,2)$

Answer 2

Check:

• $<3,3,3>+<1,0,-1>=<4,3,2>$;

• $<3,3,3>=3\,<1,1,1>$;

• $<1,0,-1>\cdot<1,1,1>=0.$

Passed.

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You were solving

$$\vec v=\vec u+\lambda \vec w$$ with $\vec u\cdot \vec w=0$.

Multiplying by $\vec w$,

$$\vec v\cdot \vec w=0+\lambda \vec w\cdot \vec w$$ gives you $\lambda=\dfrac{\vec v\cdot \vec w}{\vec w\cdot \vec w}$ and $\vec u$ follows from

$$\vec u=\vec v-\lambda \vec w.$$