MHB Find quantity of vector projection

[evinda]

Hello! (Wave)

Let $W$ be the subspace of $\mathbb{R}^3$ that is orthogonal to the vector $w_1=(-1,-1,1)$ and $p=(x,y,z)$ the projection of the vector $v=(-1,1,2)$ onto $W$. What is $7x-11y+5z$ equal to?I have thought the following:

$\text{proj}_Wv=\frac{\langle v, w_1\rangle}{\langle w_1, w_1\rangle} w_1 \Rightarrow (x,y,z)=\frac{(-1,1,2) \cdot (-1,-1,1)}{(-1,-1,1) \cdot (-1,-1,1)} (-1,-1,1) =\frac{1-1+2}{1+1+1}(-1,-1,1)=\frac{2}{3}(-1,-1,1)=\left( -\frac{2}{3},-\frac{2}{3},\frac{2}{3}\right)$

Then we get that

$$7x-11y+5z=7 \left( -\frac{2}{3}\right)-11\left( -\frac{2}{3}\right)+5\left( \frac{2}{3}\right)=6$$Have I done something wrong? Because the possible answers are $-4,-12,9,15,-14,13$.

 

[I like Serena]

Hello! (Wave)

Let $W$ be the subspace of $\mathbb{R}^3$ that is orthogonal to the vector $w_1=(-1,-1,1)$ and $p=(x,y,z)$ the projection of the vector $v=(-1,1,2)$ onto $W$. What is $7x-11y+5z$ equal to?I have thought the following:

$\text{proj}_Wv=\frac{\langle v, w_1\rangle}{\langle w_1, w_1\rangle} w_1 \Rightarrow (x,y,z)=\frac{(-1,1,2) \cdot (-1,-1,1)}{(-1,-1,1) \cdot (-1,-1,1)} (-1,-1,1) =\frac{1-1+2}{1+1+1}(-1,-1,1)=\frac{2}{3}(-1,-1,1)=\left( -\frac{2}{3},-\frac{2}{3},\frac{2}{3}\right)$

Then we get that

$$7x-11y+5z=7 \left( -\frac{2}{3}\right)-11\left( -\frac{2}{3}\right)+5\left( \frac{2}{3}\right)=6$$Have I done something wrong? Because the possible answers are $-4,-12,9,15,-14,13$.

Hey evinda!

Your projection on $W$ is actually the projection on $W^\perp$.
Instead we should have:
$$\text{proj}_Wv=v-\text{proj}_{W^\perp}v=v - \frac{\langle v, w_1\rangle}{\langle w_1, w_1\rangle} w_1$$
(Thinking)

 

[evinda]

Hey evinda!

Your projection on $W$ is actually the projection on $W^\perp$.
Instead we should have:
$$\text{proj}_Wv=v-\text{proj}_{W^\perp}v=v - \frac{\langle v, w_1\rangle}{\langle w_1, w_1\rangle} w_1$$
(Thinking)

Ah! Is it known that this formula represents the projection of $v$ onto $W$ ? (Thinking)

Using this formula, I got that $(x,y,z)=\left( -\frac{1}{3}, \frac{5}{3}, \frac{4}{3}\right)$ and thus $7x-11y+5z=-14$.

 

[I like Serena]

Ah! Is it known that this formula represents the projection of $v$ onto $W$ ? (Thinking)

Using this formula, I got that $(x,y,z)=\left( -\frac{1}{3}, \frac{5}{3}, \frac{4}{3}\right)$ and thus $7x-11y+5z=-14$.

It's indeed a known formula and it's called vector rejection.
We can see why it works in this picture:
\begin{tikzpicture}[>=stealth',rotate=10]
%preamble \usetikzlibrary{arrows}
\draw[dashed] (3,0) -- (3,4);
\draw[dashed] (0,-1) -- (0,5) node

{$w_1^\perp$};
\draw[->] (0,0) -- (5,0) node[above] {$w_1$};
\draw[ultra thick,blue,->] (0,0) -- (3,4) node[above] {$v$};
\draw[->] (3,4) -- (0,4);
\draw[thick,->] (0,0) -- node

{$v-\frac{v\cdot w_1}{w_1\cdot w_1}w_1$} (0,4);
\draw[thick,->] (0,0) -- node[below] {$\frac{v\cdot w_1}{w_1\cdot w_1}w_1$} (3,0);
\end{tikzpicture}

The vector $\frac{v\cdot w_1}{w_1\cdot w_1}w_1$ is the so called vector projection.
From the picture we can see how we can construct the vector projection on the perpendicular space.
And if $w_1$ is a vector with length 1, these formulas simplify to $(v\cdot w_1)w_1$ respectively $v-(v\cdot w_1)w_1$. (Nerd)​

 

[evinda]

It's indeed a known formula and it's called vector rejection.
We can see why it works in this picture:
\begin{tikzpicture}[>=stealth']
%preamble \usetikzlibrary{arrows}
\draw[dashed] (3,0) -- (3,4);
\draw[dashed] (0,-1) -- (0,5) node

{$w_1^\perp$};
\draw[->] (0,0) -- (5,0) node[above] {$w_1$};
\draw[ultra thick,blue,->] (0,0) -- (3,4) node[above] {$v$};
\draw[->] (3,4) -- (0,4);
\draw[thick,->] (0,0) -- node

{$v-\frac{v\cdot w_1}{w_1\cdot w_1}w_1$} (0,4);
\draw[thick,->] (0,0) -- node[below] {$\frac{v\cdot w_1}{w_1\cdot w_1}w_1$} (3,0);
\end{tikzpicture}

The vector $\frac{v\cdot w_1}{w_1\cdot w_1}w_1$ is the so called vector projection.
From the picture we can see how we can construct the vector projection on the perpendicular space.
And if $w_1$ is a vector with length 1, these formulas simplify to $(v\cdot w_1)w_1$ respectively $v-(v\cdot w_1)w_1$. (Nerd)​

Ok, thanks a lot! (Smirk)​

 

[evinda]

With the same logic, I wanted to solve the following:

Let $W$ the subspace of $\mathbb{R}^3$ that is spanned by $w_1=(1,-1,0)$, $w_2=(1,1,2)$ and $p=(x,y,z)$ the projection of the vector $v=(-1,1,2)$ onto $W$. What is the number $2x+7y+3z$ equal to?

Using the formula, I got that

$$(x,y,z)=(-1,1,2)- \left( \frac{(-1,1,2) \cdot (1,-1,0)}{(1,-1,0) \cdot (1,-1,0)}(1,-1,0)+\frac{(-1,1,2) \cdot (1,1,2)}{(1,1,2) \cdot (1,1,2)}(1,1,2)\right)=\left( -\frac{2}{3},-\frac{2}{3}, \frac{2}{3}\right)$$

Then $2x+7y+3z=-4$. But this isn't a possible answer... (Worried)

Have I done something wrong? (Thinking)

 

[I like Serena]

With the same logic, I wanted to solve the following:

Let $W$ the subspace of $\mathbb{R}^3$ that is spanned by $w_1=(1,-1,0)$, $w_2=(1,1,2)$ and $p=(x,y,z)$ the projection of the vector $v=(-1,1,2)$ onto $W$. What is the number $2x+7y+3z$ equal to?

Using the formula, I got that

$$(x,y,z)=(-1,1,2)- \left( \frac{(-1,1,2) \cdot (1,-1,0)}{(1,-1,0) \cdot (1,-1,0)}(1,-1,0)+\frac{(-1,1,2) \cdot (1,1,2)}{(1,1,2) \cdot (1,1,2)}(1,1,2)\right)=\left( -\frac{2}{3},-\frac{2}{3}, \frac{2}{3}\right)$$

Then $2x+7y+3z=-4$. But this isn't a possible answer... (Worried)

Have I done something wrong? (Thinking)

This time round we have vectors that span $W$ instead of being perpendicular to $W$.
So we need the vector projection formula instead of the vector rejection formula.

Alternatively we can find the vector $n$ that is perpendicular to both $w_1$ and $w_2$, and then we can use the vector rejection formula. (Thinking)

 

[evinda]

This time round we have vectors that span $W$ instead of being perpendicular to $W$.
So we need the vector projection formula instead of the vector rejection formula.

Alternatively we can find the vector $n$ that is perpendicular to both $w_1$ and $w_2$, and then we can use the vector rejection formula. (Thinking)

Ah ok... So we have that $(x,y,z)=\frac{(-1,1,2) \cdot (1,-1,0)}{(1,-1,0) \cdot (1,-1,0)}(1,-1,0)+\frac{(-1,1,2) \cdot (1,1,2)}{(1,1,2) \cdot (1,1,2)} \cdot (1,1,2)=\left( -\frac{1}{3}, \frac{5}{3}, \frac{4}{3}\right)$ and thus $2x+7y+3z=15$, right? (Thinking)

 

[I like Serena]

Ah ok... So we have that $(x,y,z)=\frac{(-1,1,2) \cdot (1,-1,0)}{(1,-1,0) \cdot (1,-1,0)}(1,-1,0)+\frac{(-1,1,2) \cdot (1,1,2)}{(1,1,2) \cdot (1,1,2)} \cdot (1,1,2)=\left( -\frac{1}{3}, \frac{5}{3}, \frac{4}{3}\right)$ and thus $2x+7y+3z=15$, right? (Thinking)

Yep. (Nod)

 

[evinda]

Yep. (Nod)

Nice... thank you (Smirk)