Struggling with this question. Find two vectors $v_1$ and $v_2$ such that when added equal $(0, 4, 0)$. $v_1$ is parallel to $u(-2, 4, -2)$ and $v_2$ is perpendicular to $u$. Not sure how to start.
$\endgroup$5 Answers
$\begingroup$$\mathbf{v_1}$ is parallel to $\mathbf{u} = (-2,4,-2)$ means that $\mathbf{v_1}$ has the same direction. So, it's just a scalar multiple of $\mathbf{u}$: $\mathbf{v_1}=k\mathbf{u}$ for some scalar $k$.
If $\mathbf{v_2}$ is perpendicular to $\mathbf{u}$ then that means, among other things, that the dot product between the two is $0$. Say $\mathbf{v_2} = (a,b,c)$. Then we know that $\mathbf{v_2}\cdot\mathbf{u} = (a,b,c) \cdot (-2,4,-2) = 0 $.
Finally, you know that $\mathbf{v_1} + \mathbf{v_2} = (0,4,0)$.
This is how you start these problems. If nothing immediately looks obvious or familiar, write out things you know, try to decipher what they mean, and see if you see anything once things are a bit simplified.
$\endgroup$ $\begingroup$Denote by $a \cdot b$ the scalar product of two vectors.
Since $v_1$ must be parallel to $u$, write $v_1=\lambda u$ with $\lambda > 0$.
Now, $$16=(0,4,0) \cdot u = (v_1+v_2)\cdot u = v_1 \cdot u + v_2 \cdot u = \lambda |u|^2 + 0 = 12 \lambda $$ so $\lambda = \frac43$.
Then necessarily, $v_1= \frac43 u$ and $v_2 = (0,4,0) - v_1$, and I leave you computations.
$\endgroup$ $\begingroup$If you have a vector $\underline{e}$ of unit length, then for any vector $\underline{v}$
$$\underline{v}=\underbrace{\langle \underline{v},\underline{e}\rangle \underline{e}}_{\text{this is parallel to } \underline{e}}+\underbrace{\underline{v}-\langle \underline{v},\underline{e}\rangle \underline{e}}_{\text{this is perpendicular to } \underline{e}}$$
Where $\langle\underline{x},\underline{y}\rangle$ is the scalar product.
The parallelness is trivial (scalar times $\underline{e}$, the other is easy, just use linearity and $\langle\underline{e},\underline{e}\rangle=1$: $$ \langle\underline{e},\underline{v}-\langle \underline{v},\underline{e}\rangle \underline{e}\rangle=\langle\underline{e},\underline{v}\rangle-\langle\underline{e},\underline{e}\langle\underline{e},\underline{v}\rangle\rangle=\langle\underline{e},\underline{v}\rangle-\langle\underline{e},\underline{v}\rangle\langle\underline{e},\underline{e}\rangle=0 $$
In this case $\underline{v}=(0,4,0)$ and $\underline{e}=\frac{1}{\sqrt{6}}(-1,2,-1)$
$\endgroup$ $\begingroup$You have $v_1+v_2=\left(\begin{matrix}0\\4\\0\end {matrix}\right)$, $v_1=\lambda\left(\begin{matrix}-2\\4\\-2\end {matrix}\right)$, and $v_2\cdot\left(\begin{matrix}-2\\4\\-2\end {matrix}\right)=0$
$$v_2=\left(\begin{matrix}0\\4\\0\end {matrix}\right)-\lambda\left(\begin{matrix}-2\\4\\-2\end {matrix}\right)\Rightarrow\left(\begin{matrix}2\lambda\\4-4\lambda\\2\lambda\end {matrix}\right)\cdot\left(\begin{matrix}-2\\4\\-2\end {matrix}\right)=0$$
Hence $\lambda=\frac 23$ so you can deduce $v_1$ and $v_2$
$\endgroup$ $\begingroup$$v_1+v_2=(0,4,0),$ $v_2\cdot{(-2,4,-2)}=0, v_1\times (-2,4,-2)=0 $ where $\times$ cross product and $\cdot$ dot product. Solving a system of equations we get $v_1=(-4/3,8/3,-4/3)$ and $v_2=(4/3,4/3,4/3)$.
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