I can"t figure this out. How the book explains to do this problem doesn"t make sense to me. I"m hoping someone can simplify this so I can understand how to do this step by step.

Let $L$ be the line in $ asiilaq.netbbR^3$ that consists of all scalar multiples of $$ left< eginarrayccc 2\ 1\ 2\ endarray ight> $$ Find the reflection of the vector

$$ left< eginarrayccc 1\ 4\ 1\ endarray ight> $$ onto $L$




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Let $L$ be the line spanned by some non-zero vector $ asiilaq.netbfv in asiilaq.netbbR^3$, so that $L = a asiilaq.netbfv mid a in asiilaq.netbbR$ is the space of all scalar multiples of $ asiilaq.netbfv$. Then the orthogonal projection of a vector $ asiilaq.netbfx in asiilaq.netbbR^3$ onto the line $L$ can be computed as$$ operatornameProj_L( asiilaq.netbfx) = frac asiilaq.netbfv cdot asiilaq.netbfx asiilaq.netbfvcdot asiilaq.netbfv asiilaq.netbfv.$$So, in this case, we have $$ asiilaq.netbfv = eginpmatrix2\1\2endpmatrix, quad asiilaq.netbfx = eginpmatrix1\4\1endpmatrix,$$so that$$ asiilaq.netbfv cdot asiilaq.netbfx = 2 cdot 1 + 1 cdot 4 + 2 cdot 1 = 8, quad asiilaq.netbfv cdot asiilaq.netbfv = 2^2 + 1^2 + 2^2 = 9,$$and hence$$ operatornameProj_L( asiilaq.netbfx) = frac89eginpmatrix2\1\2endpmatrix.$$

Now, you probably wanted to compute the orthogonal projection of some vector $ asiilaq.netbfx$ onto the line $L$ spanned by some non-zero vector $ asiilaq.netbfv$. But what if you did want to compute the reflection of $ asiilaq.netbfx$ in the line $L$? What would this mean? Well, in general, suppose that $S$ is a subspace of $ asiilaq.netbbR^3$ (e.g., a line through the origin or a plane through the origin), so that for any vector $ asiilaq.netbfx in asiilaq.netbbR^3$, we have$$ asiilaq.netbfx = operatornameProj_S( asiilaq.netbfx) + operatornameProj_S^perp( asiilaq.netbfx),$$where $operatornameProj_S( asiilaq.netbfx)$ is the orthogonal projection of $ asiilaq.netbfx$ onto $S$ and $operatornameProj_S^perp( asiilaq.netbfx)$ is the orthogonal projection of $ asiilaq.netbfx$ onto the orthogonal complement$$ S^perp = \tasiilaq.netbfy in asiilaq.netbbR^3 mid extfor every $ asiilaq.netbfz in S$, ; asiilaq.netbfy cdot asiilaq.netbfz = 0$$of $S$; in particular, observe that$$ operatornameProj_S^perp( asiilaq.netbfx) = asiilaq.netbfx - operatornameProj_S( asiilaq.netbfx).$$Then, geometrically, the reflection $operatornameRefl_S( asiilaq.netbfx)$ of $ asiilaq.netbfx$ in $S$ is given by fixing the component of $ asiilaq.netbfx$ in $S$ and flipping the direction of the component of $ asiilaq.netbfx$ in $S^perp$, i.e.,$$ operatornameRefl_S( asiilaq.netbfx) = operatornameProj_S( asiilaq.netbfx) - operatornameProj_S^perp( asiilaq.netbfx) = operatornameProj_S( asiilaq.netbfx) - left( asiilaq.netbfx - operatornameProj_S( asiilaq.netbfx) ight) = 2operatornameProj_S( asiilaq.netbfx) - asiilaq.netbfx.$$So, suppose that $L$ is the line spanned by some non-zero vector $ asiilaq.netbfv in asiilaq.netbbR^3$.

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On the one hand, the reflection of $ asiilaq.netbfx$ in $L$ is given by$$ operatornameRefl_L( asiilaq.netbfx) = 2operatornameProj_L( asiilaq.netbfx) - asiilaq.netbfx = 2frac asiilaq.netbfv cdot asiilaq.netbfx asiilaq.netbfvcdot asiilaq.netbfv asiilaq.netbfv - asiilaq.netbfx,$$which in your case yields$$ operatornameRefl_L( asiilaq.netbfx) = 2 cdot frac89eginpmatrix2\1\2endpmatrix - eginpmatrix1\4\1endpmatrix = frac19eginpmatrix23\-20\23endpmatrix.$$On the other hand, the reflection of $ asiilaq.netbfx$ in the plane$$ L^perp = \tasiilaq.netbfy in asiilaq.netbbR^3 mid asiilaq.netbfv cdot asiilaq.netbfy = 0$$with normal vector $ asiilaq.netbfv$ is given by$$ operatornameRefl_L^perp( asiilaq.netbfx) = 2operatornameProj_L^perp( asiilaq.netbfx) - asiilaq.netbfx = 2left( asiilaq.netbfx - operatornameProj_L( asiilaq.netbfx) ight) - asiilaq.netbfx = asiilaq.netbfx - 2operatornameProj_L( asiilaq.netbfx) = asiilaq.netbfx - 2frac asiilaq.netbfv cdot asiilaq.netbfx asiilaq.netbfvcdot asiilaq.netbfv asiilaq.netbfv,$$which in your case yields$$ operatornameRefl_L^perp( asiilaq.netbfx) = eginpmatrix1\4\1endpmatrix - 2 cdot frac89eginpmatrix2\1\2endpmatrix = -frac19eginpmatrix23\-20\23endpmatrix.$$One last cultural note: the reflection $operatornameRefl_L^perp( asiilaq.netbfx)$ of $ asiilaq.netbfx$ in the plane $L^perp$ with normal vector $ asiilaq.netbfv$ is better known in more advanced contexts by another name, namely as the image$$ H_ asiilaq.netbfv( asiilaq.netbfx) := operatornameRefl_L^perp( asiilaq.netbfx) = asiilaq.netbfx - 2frac asiilaq.netbfv cdot asiilaq.netbfx asiilaq.netbfvcdot asiilaq.netbfv asiilaq.netbfv$$of $ asiilaq.netbfx$ under the Householder transformation $H_ asiilaq.netbfv$ corresponding to $ asiilaq.netbfv$. So, in your case, as we just saw,$$ H_ asiilaq.netbfv( asiilaq.netbfx) = -frac19eginpmatrix23\-20\23endpmatrix.$$