I can"t number this out. Just how the publication explains to execute this problem doesn"t make feeling to me. I"m hope someone have the right to simplify this for this reason I can understand just how to carry out this step by step.

Let $L$ be the heat in $\tasiilaq.netbbR^3$ that consists of all scalar multiples the $$\left< \beginarrayccc 2\\ 1\\ 2\\ \endarray\right>$$ uncover the enjoy of the vector

$$\left< \beginarrayccc 1\\ 4\\ 1\\ \endarray\right>$$ ~ above $L$

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Let $L$ be the line extended by some non-zero vector $\tasiilaq.netbfv \in \tasiilaq.netbbR^3$, so the $L = \a\tasiilaq.netbfv \mid a \in \tasiilaq.netbbR\$ is the space of every scalar multiples of $\tasiilaq.netbfv$. Climate the orthogonal projection of a vector $\tasiilaq.netbfx \in \tasiilaq.netbbR^3$ ~ above the line $L$ deserve to be computed as$$\operatornameProj_L(\tasiilaq.netbfx) = \frac\tasiilaq.netbfv \cdot \tasiilaq.netbfx\tasiilaq.netbfv\cdot\tasiilaq.netbfv \tasiilaq.netbfv.$$So, in this case, we have $$\tasiilaq.netbfv = \beginpmatrix2\\1\\2\endpmatrix, \quad \tasiilaq.netbfx = \beginpmatrix1\\4\\1\endpmatrix,$$so that$$\tasiilaq.netbfv \cdot \tasiilaq.netbfx = 2 \cdot 1 + 1 \cdot 4 + 2 \cdot 1 = 8, \quad \tasiilaq.netbfv \cdot \tasiilaq.netbfv = 2^2 + 1^2 + 2^2 = 9,$$and hence$$\operatornameProj_L(\tasiilaq.netbfx) = \frac89\beginpmatrix2\\1\\2\endpmatrix.$$

Now, you more than likely wanted to compute the orthogonal estimate of part vector $\tasiilaq.netbfx$ top top the line $L$ covered by some non-zero vector $\tasiilaq.netbfv$. Yet what if girlfriend did want to compute the reflection of $\tasiilaq.netbfx$ in the heat $L$? What would certainly this mean? Well, in general, expect that $S$ is a subspace that $\tasiilaq.netbbR^3$ (e.g., a line v the origin or a plane through the origin), so that for any kind of vector $\tasiilaq.netbfx \in \tasiilaq.netbbR^3$, us have$$\tasiilaq.netbfx = \operatornameProj_S(\tasiilaq.netbfx) + \operatornameProj_S^\perp(\tasiilaq.netbfx),$$where $\operatornameProj_S(\tasiilaq.netbfx)$ is the orthogonal estimate of $\tasiilaq.netbfx$ onto $S$ and also $\operatornameProj_S^\perp(\tasiilaq.netbfx)$ is the orthogonal forecast of $\tasiilaq.netbfx$ ~ above the orthogonal complement$$S^\perp = \\tasiilaq.netbfy \in \tasiilaq.netbbR^3 \mid \textfor every \tasiilaq.netbfz \in S, \; \tasiilaq.netbfy \cdot \tasiilaq.netbfz = 0\$$of $S$; in particular, watch that$$\operatornameProj_S^\perp(\tasiilaq.netbfx) = \tasiilaq.netbfx - \operatornameProj_S(\tasiilaq.netbfx).$$Then, geometrically, the reflection $\operatornameRefl_S(\tasiilaq.netbfx)$ the $\tasiilaq.netbfx$ in $S$ is offered by addressing the component of $\tasiilaq.netbfx$ in $S$ and flipping the direction the the ingredient of $\tasiilaq.netbfx$ in $S^\perp$, i.e.,$$\operatornameRefl_S(\tasiilaq.netbfx) = \operatornameProj_S(\tasiilaq.netbfx) - \operatornameProj_S^\perp(\tasiilaq.netbfx) = \operatornameProj_S(\tasiilaq.netbfx) - \left( \tasiilaq.netbfx - \operatornameProj_S(\tasiilaq.netbfx)\right) = 2\operatornameProj_S(\tasiilaq.netbfx) - \tasiilaq.netbfx.$$So, expect that $L$ is the line extended by part non-zero vector $\tasiilaq.netbfv \in \tasiilaq.netbbR^3$.

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On the one hand, the reflection of $\tasiilaq.netbfx$ in $L$ is provided by$$\operatornameRefl_L(\tasiilaq.netbfx) = 2\operatornameProj_L(\tasiilaq.netbfx) - \tasiilaq.netbfx = 2\frac\tasiilaq.netbfv \cdot \tasiilaq.netbfx\tasiilaq.netbfv\cdot\tasiilaq.netbfv \tasiilaq.netbfv - \tasiilaq.netbfx,$$which in your case yields$$\operatornameRefl_L(\tasiilaq.netbfx) = 2 \cdot \frac89\beginpmatrix2\\1\\2\endpmatrix - \beginpmatrix1\\4\\1\endpmatrix = \frac19\beginpmatrix23\\-20\\23\endpmatrix.$$On the other hand, the reflection of $\tasiilaq.netbfx$ in the plane$$L^\perp = \\tasiilaq.netbfy \in \tasiilaq.netbbR^3 \mid \tasiilaq.netbfv \cdot \tasiilaq.netbfy = 0\$$with common vector $\tasiilaq.netbfv$ is offered by$$\operatornameRefl_L^\perp(\tasiilaq.netbfx) = 2\operatornameProj_L^\perp(\tasiilaq.netbfx) - \tasiilaq.netbfx = 2\left(\tasiilaq.netbfx - \operatornameProj_L(\tasiilaq.netbfx)\right) - \tasiilaq.netbfx = \tasiilaq.netbfx - 2\operatornameProj_L(\tasiilaq.netbfx) = \tasiilaq.netbfx - 2\frac\tasiilaq.netbfv \cdot \tasiilaq.netbfx\tasiilaq.netbfv\cdot\tasiilaq.netbfv \tasiilaq.netbfv,$$which in your case yields$$\operatornameRefl_L^\perp(\tasiilaq.netbfx) = \beginpmatrix1\\4\\1\endpmatrix - 2 \cdot \frac89\beginpmatrix2\\1\\2\endpmatrix = -\frac19\beginpmatrix23\\-20\\23\endpmatrix.$$One last cultural note: the have fun $\operatornameRefl_L^\perp(\tasiilaq.netbfx)$ of $\tasiilaq.netbfx$ in the airplane $L^\perp$ through normal vector $\tasiilaq.netbfv$ is much better known in an ext advanced contexts by one more name, namely as the image$$H_\tasiilaq.netbfv(\tasiilaq.netbfx) := \operatornameRefl_L^\perp(\tasiilaq.netbfx) = \tasiilaq.netbfx - 2\frac\tasiilaq.netbfv \cdot \tasiilaq.netbfx\tasiilaq.netbfv\cdot\tasiilaq.netbfv \tasiilaq.netbfv$$of $\tasiilaq.netbfx$ under the Householder transformation $H_\tasiilaq.netbfv$ matching to $\tasiilaq.netbfv$. So, in your case, as we simply saw,$$H_\tasiilaq.netbfv(\tasiilaq.netbfx) = -\frac19\beginpmatrix23\\-20\\23\endpmatrix.$$