heureka

In triangle ABC (see Figure 1.4), \(\angle BCA = 90^\circ\) , and D is the foot of
the perpendicular line segment from C to segment AB. Given that |AB| = x and
\(\angle A = \theta\), express all the lengths of the segments in Figure 1.4 in terms of x and \(\theta\).

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{|CB|} &=& \mathbf{x\cdot \sin(\theta)} \\ (2) & \mathbf{|CA|} &=& \mathbf{x\cdot \cos(\theta)} \\\\ & \sin(\theta) &=& \dfrac{|CD|}{|CA|} \\ & |CD| &=& |CA|\cdot \sin(\theta) \\ (3)& \mathbf{|CD|} &=& \mathbf{x\cdot \cos(\theta)\cdot \sin(\theta) } \\\\ & \cos(\theta) &=& \dfrac{|AD|}{|CA|} \\ & |AD| &=& |CA|\cdot \cos(\theta) \\ & |AD| &=& x\cdot \cos(\theta)\cdot \cos(\theta) \\ (4)& \mathbf{|AD|} &=& \mathbf{x\cdot \cos^2(\theta)} \\\\ & |DB| &=& x-|AD| \\ & |DB| &=& x-x\cdot \cos^2(\theta) \\ & |DB| &=& x\cdot\left(1-\cos^2(\theta)\right) \quad | \quad \sin^2(\theta)+cos^2(\theta)=1 \\ (5)& \mathbf{|DB|} &=& \mathbf{x\cdot\sin^2(\theta)} \\ \hline \end{array}\)

Consider the matrices

\(\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \mathbf{B} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 &3 \\ 1 & 3 & 5 \end{pmatrix}, \mathbf{C} = \begin{pmatrix} 1 & -2 & 1 \\ 2 & -4 &2 \\ 3 & -6 & 3 \end{pmatrix}, \mathbf{D} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 1 & 0 \end{pmatrix}.\)

\(\boxed{\text{Any matrix }\mathbf{A}\text{ can be inverted if the following applies: } \mathbf{det(A)\ne 0} } \)

\(\begin{array}{|rcll|} \hline \det(A) &=& \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=& 1\cdot 1 \cdot 1 +0\cdot 0 \cdot 0 +0\cdot 0 \cdot 0 -0\cdot 1 \cdot 0 -1\cdot 0 \cdot 0 -0\cdot 0 \cdot 1 \\ &=& 1+0+0-0-0-0 \\ &=& \mathbf{1} \quad | \quad \mathbf{A} \text{ is invertible } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \det(B) &=& \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 &3 \\ 1 & 3 & 5 \end{vmatrix} \\ &=& 1\cdot 2 \cdot 5 +1\cdot 1 \cdot 3 +1\cdot 3 \cdot 1 -1\cdot 2 \cdot 1 -1\cdot 3 \cdot 3 -1\cdot 1 \cdot 5 \\ &=& 10+3+3-2-9-5 \\ &=& \mathbf{0} \quad | \quad \mathbf{B} \text{ is not invertible } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \det(C) &=& \begin{vmatrix} 1 & -2 & 1 \\ 2 & -4 &2 \\ 3 & -6 & 3 \end{vmatrix} \\ &=& 1\cdot (-4) \cdot 3 +3\cdot (-2) \cdot 2 +2\cdot (-6) \cdot 1 -3\cdot (-4) \cdot 1 -1\cdot (-6) \cdot 2 -2\cdot (-2) \cdot 3 \\ &=& -12-12-12+12+12+12 \\ &=& \mathbf{0} \quad | \quad \mathbf{C} \text{ is not invertible } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \det(D) &=& \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 1 & 1 & 0 \end{vmatrix} \\ &=& 1\cdot 1 \cdot 0 +1\cdot 0 \cdot 1 +0\cdot 1 \cdot 1 -1\cdot 1 \cdot 1 -1\cdot 1 \cdot 1 -0\cdot 0 \cdot 0 \\ &=& 0+0+0-1-1-0 \\ &=& \mathbf{-2} \quad | \quad \mathbf{D} \text{ is invertible } \\ \hline \end{array}\)

Find the distance between Q = (3, -7, -1) and the line through A = (1, 1, 2) and B = (2, 3, 4).
This distance is equal to \(\dfrac{\sqrt{d}}{3}\) for some integer d.
What is d?

line through A = (1, 1, 2) and B = (2, 3, 4):

\(\small{ \begin{array}{|rclrcl|} \hline \vec{x} &=& \vec{A}+t(\vec{B}-\vec{A}) & \vec{r} &=& \vec{B}-\vec{A} \quad | \quad \vec{A} = (1, 1, 2)\quad \vec{B} = (2, 3, 4)\\ &&& \vec{r} &=& \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}-\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ &&&\vec{r} &=& \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\ \mathbf{\vec{x}} &=& \mathbf{\vec{A}+t\vec{r}} \\ \hline (\vec{x}-\vec{Q})\cdot \vec{r} &=& 0 \quad | \quad \vec{PQ} \text{ is } \perp \text{ to } \vec{r} \\ (\mathbf{\vec{A}+t\vec{r}}-\vec{Q})\cdot \vec{r} &=& 0 \\ (\vec{A}-\vec{Q} + t\vec{r})\cdot \vec{r} &=& 0 & \vec{A}-\vec{Q}&=& \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}-\begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix} \quad | \quad \vec{Q} = (3, -7, -1)\\ & & & \vec{A}-\vec{Q}&=& \begin{pmatrix} -2 \\ 8 \\ 3 \end{pmatrix} \\\\ \left[\begin{pmatrix} -2 \\ 8 \\ 3 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \right]\cdot\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} &=& 0 \\\\ \left[\begin{pmatrix} -2+ t \\ 8+2t \\ 3+2t \end{pmatrix} \right]\cdot\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} &=& 0 \\\\ (-2+ t)\cdot 1 + (8+2t)\cdot 2+ (3+2t)\cdot 2 &=& 0 \\ -2+t+16+4t+6+4t &=& 0 \\ 9t+20 &=& 0 \\ \mathbf{t} &=& \mathbf{ -\dfrac{20}{9} } \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline \mathbf{\vec{P}} &=& \vec{A}+t\vec{r} \\ &=& \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}-\mathbf{ \dfrac{20}{9} }\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\\\ &=& \begin{pmatrix} 1-\dfrac{20}{9} \\ 1-\dfrac{40}{9} \\ 2-\dfrac{20}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{ -\dfrac{11}{9}} \\ \mathbf{-\dfrac{31}{9}} \\ \mathbf{ -\dfrac{22}{9}} \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\vec{PQ}} &=& \vec{Q}- \vec{P} \\ &=& \begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix}- \begin{pmatrix} \mathbf{ -\dfrac{11}{9}} \\ \mathbf{-\dfrac{31}{9}} \\ \mathbf{ -\dfrac{22}{9}} \end{pmatrix} \\\\ &=& \begin{pmatrix} 3 \\ -7 \\ -1 \end{pmatrix}+ \begin{pmatrix} \dfrac{11}{9} \\ \dfrac{31}{9} \\ \dfrac{22}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} 3 +\dfrac{11}{9} \\ -7+\dfrac{31}{9} \\ -1+\dfrac{22}{9} \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{ \dfrac{38}{9}} \\ \mathbf{-\dfrac{32}{9}} \\ \mathbf{ \dfrac{13}{9}} \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline |\vec{PQ}| &=& \sqrt{ \mathbf{ \left( \dfrac{38}{9} \right)^2} +\mathbf{\left(-\dfrac{32}{9}\right)^2 } +\mathbf{\left( \dfrac{13}{9}\right)^2 } } \\ &=& \sqrt{ \dfrac{38^2}{9^2} + \dfrac{32^2}{9^2} + \dfrac{13^2}{9^2} } \\\\ &=& \sqrt{ \dfrac{1}{9}\left( \dfrac{38^2}{9} + \dfrac{32^2}{9} + \dfrac{13^2}{9} \right) } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{38^2}{9} + \dfrac{32^2}{9} + \dfrac{13^2}{9} } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{1}{9}\left( 38^2+32^2+13^2 \right) } \\\\ &=& \dfrac{1}{3}\sqrt{ \dfrac{2637}{9} } \\\\ &=& \dfrac{1}{3}\sqrt{293} \\\\ &=& \mathbf{ \dfrac{\sqrt{293}}{3} } \quad | \quad \dfrac{\sqrt{d}}{3} \\ \\ \mathbf{d} &=& \mathbf{293}\\ \hline \end{array}\)

Consider the line through points A = (1, 1, 2) and B = (2, 3, 4),
and let P be the foot of the perpendicular from Q = (3, -7, -1) to this line,
as in the picture below:

Then \(\vec{AP}\) is the projection of \(\vec{AQ}\) onto a vector
\(\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}\)
such that \(u_1 + u_2 + u_3 = 5\).

What is u?

\(\begin{array}{|rcll|} \hline \vec{u} &=& \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} \\ &=& \vec{B} - \vec{A} \\ &=& \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ &=& \begin{pmatrix} 2-1 \\ 3-1 \\ 4-2 \end{pmatrix} \\ &=& \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \\\\ && u_1 + u_2 + u_3 \\ &=& 1+2+2 \\ &=& 5 \\ \hline \end{array}\)

Compute  \(\large{\left(1 + i\right)^4}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\left(1 + i\right)^4} \\ &=& \Big(~\left(1 + i\right)^2~\Big)^2 \quad &| \quad \left(1 + i\right)^2 = 1+2i+i^2 \\ &=& \Big(~1+2i+i^2~\Big)^2 \quad &| \quad i^2=-1 \\ &=& \left(~1+2i-1~\right)^2 \\ &=& \left( 2i \right)^2 \\ &=&4i^2\quad &| \quad i^2=-1 \\ &=&4(-1) \\ &=& \mathbf{-4} \\ \hline \end{array}\)

The graph of the equation \(y = a|x-b| + c\) has two x-intercepts at \(x=-3\) and \(x=7\),and one y-intercept at \(y=5\).
What is the product \(abc\) ?

\(\begin{array}{|lrclrcl|} \hline \mathbf{x-\text{intercepts}\quad y=0:} & 0 &=& a|x-b| + c \\ & \mathbf{|x-b|}&=& \mathbf{-\dfrac{c}{a}} \\\\ & x&=& -3 & x&=&7 \\ & |-3-b|&=& -\dfrac{c}{a} & |7-b|&=& -\dfrac{c}{a} \\ & |-(3+b)|&=& -\dfrac{c}{a} & \left(7-b\right)^2&=& \left(-\dfrac{c}{a}\right)^2 \quad(1) \\ & |3+b|&=& -\dfrac{c}{a} \\ & \left(3+b\right)^2&=& \left(-\dfrac{c}{a}\right)^2 \quad(2) \\\\ (1)=(2): & \left(3+b\right)^2&=& \left(7-b\right)^2 \\ & 9+6b+b^2 &=&49-14b+b^2 \\ & 9+6b &=&49-14b \\ &20b &=& 40 \\ & \mathbf{b} &=& \mathbf{2} \qquad |b|=2 \\\\ & |3+b|&=& -\dfrac{c}{a} \\ & |3+2|&=& -\dfrac{c}{a} \\ & 5 &=& -\dfrac{c}{a} \\ & 5a &=& -c \\ & \mathbf{ a } &=& \mathbf{-\dfrac{c}{5}} \\ \hline \mathbf{y-\text{intercepts}\quad x=0:} & y &=& a|0-b| + c \\ & y &=& a|b| + c \\ & \mathbf{y} &=& \mathbf{2a + c} \\\\ & y &=& 5 \\ & 5 &=& 2a+c \quad | \quad a =-\dfrac{c}{5} \\ & 5 &=& 2\left(-\dfrac{c}{5}\right)+c \\ & 5 &=& -\dfrac{2}{5} c+c \\ & 5 &=& \dfrac{3}{5} c \\ & \mathbf{ c } &=& \mathbf{ \dfrac{25}{3}} \\\\ & a &=& -\dfrac{c}{5} \\ & a &=& -\dfrac{\dfrac{25}{3}}{5} \\ & \mathbf{ a} &=& \mathbf{- \dfrac{5}{3}} \\ \hline \end{array}\)

\(\boxed{\mathbf{y = \left( - \dfrac{5}{3}\right)|x-2| + \dfrac{25}{3}}} \\\)

\(\begin{array}{|rcll|} \hline abc &=& \left( - \dfrac{5}{3}\right)\cdot 2 \cdot \dfrac{25}{3} \\ \mathbf{abc} &=& \mathbf{-\left( \dfrac{250}{9}\right)} \\ \hline \end{array}\)

Let \(a, b, c, d, e, f\) be nonnegative real numbers such that \(a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6\) and \(ab + cd + ef = 3\).
What is the maximum value of \(a+b+c+d+e+f\) ?

The Cauchy–Schwarz inequality states that for all vectors  \(\mathbf{u}\) and  \(\mathbf{v}\) of an inner product space it is true that
\({\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle },\)
where \({\displaystyle \langle \cdot ,\cdot \rangle }\) is the inner product.

\(\text{Let $\vec{u} = \begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix}$ } \\ \text{Let $\vec{v} = \begin{pmatrix} 1 \\1\\1 \end{pmatrix}$ } \)

\(\begin{array}{|rcll|} \hline \langle \mathbf {u} ,\mathbf {v} \rangle &=& \begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& a+b+c+d+e+f \\\\ \langle \mathbf {u} ,\mathbf {u} \rangle &=& \begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix}\begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix} \\ &=& a^2+b^2+c^2+d^2+e^2+f^2+2(ab + cd + ef) \\ &=& 6+2(3) \\ &=&\mathbf{ 12 } \\\\ \langle \mathbf {v} ,\mathbf {v} \rangle &=& \begin{pmatrix} 1 \\1\\1 \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& 1^2+1^2+1^2 \\ &=& \mathbf{ 3 } \\\\ \hline |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2} &\leq& \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle \\ (a+b+c+d+e+f)^2 &\le& 12\cdot 3 \\ (a+b+c+d+e+f)^2 &\le& 36 \\ \mathbf{ a+b+c+d+e+f } & \mathbf{\le} & \mathbf{6} \\ \hline \end{array}\)

The maximum value of \(a+b+c+d+e+f\) is \(\mathbf{6}\).

Let a and b be positive real numbers such that \(a + 2b = 1\).  
Find the minimum value of  \(\dfrac{2}{a} + \dfrac{1}{b}\).

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{2}{a} + \dfrac{1}{b}} \\\\ &=& \dfrac{a+2b}{ab} \quad | \quad a + 2b = 1 \\\\ &=& \mathbf{\dfrac{1}{ab}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \large{GM} &\large{\leq}& \large{AM} \\\\ \sqrt{a(2b)} & \leq & \dfrac{ a+2b}{2} \quad | \quad a + 2b = 1 \\ \sqrt{a(2b)} & \leq & \dfrac{ 1}{2} \quad | \quad \text{square both sides} \\ 2ab & \leq & \dfrac{ 1}{4} \quad | \quad :2 \\ ab & \leq & \dfrac{ 1}{8} \\ && \boxed{ \text{If the sides of the inequality are either both positive or both negative, applies:} \\ \text{If the reciprocal value is formed on both sides of an inequality} \\ \text{the inequality sign turns around} } \\ \dfrac{1}{ab} & \geq & 8 \\ \hline \end{array}\)

The minimum value of  \(\mathbf{\dfrac{1}{ab}}=\dfrac{2}{a} + \dfrac{1}{b}\) is 8  

The equation of the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2) can be written as ax + by + cz = 13. 

What is the ordered triple   (a, b, c)? 

Formula:  \(\vec{x} = \vec{C} + r(\vec{A}-\vec{C}) +s(\vec{B}-\vec{C})\)

\(\begin{array}{|lrcll|} \hline & \mathbf{\vec{x}} &=& \mathbf{ \vec{C} + r(\vec{A}-\vec{C}) +s(\vec{B}-\vec{C}) } \\ & \begin{pmatrix} x\\ y\\ z \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r\left(\begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) +s\left(\begin{pmatrix} -2\\ 1\\ 4 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) \\ & \begin{pmatrix} x\\ y\\ z \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r \begin{pmatrix} 0\\ -3\\ -1 \end{pmatrix} +s \begin{pmatrix} -4\\ 0\\ 2 \end{pmatrix} \\ \hline (1) & x &=& 1-4s \\ & 4s &=& 1-x \\ & \mathbf{s} &=& \mathbf{\dfrac{1}{4}(1-x)} \\\\ (2) & y &=& 1-3r \\ & 3r &=& 1-y \\ & \mathbf{r} &=& \mathbf{\dfrac{1}{3}(1-y)} \\\\ (3) & z &=& 2-r+2s \\ & z &=& 2-\dfrac{1}{3}(1-y)+\dfrac{2}{4}(1-x) \\ & z &=& 2-\dfrac{1}{3}(1-y)+\dfrac{1}{2}(1-x) \quad | \quad \cdot 6 \\ & 6z &=& 12-\dfrac{6}{3}(1-y)+\dfrac{6}{2}(1-x) \\ & 6z &=& 12-2(1-y)+3(1-x) \\ & 6z &=& 12-2+2y+3-3x \\ & \mathbf{3x-2y+6z } &=& \mathbf{13} \\ \hline \end{array}\)

\((a,\ b,\ c) = (3,\ -2,\ 6)\)

long poly div help

\(\begin{array}{|rcll|} \hline (k^3-4k^2-30k-18) : (k+3) = \mathbf{k^2-7k-9+\dfrac{9}{k+3}} \\ \hline \end{array}\)