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Consider points \(A, B\) and \(P\) in the picture below, with \(\angle APB\) a right angle:


Then \(\overrightarrow{AP}\) is the projection of \(\overrightarrow{AB}\) onto some vector  u , with \(\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}\)
If \(u_1 + u_2 = 3\), what is u?

 Feb 9, 2019
 #1
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Consider

\(\text{points $A$,$B$ and $P$ in the picture below, with $\angle APB$ a right angle:}\)

\(\text{Then $\overrightarrow{AP}$ is the projection of $\overrightarrow{AB}$ onto some vector $u$ , } \\ \text{with $\mathbf{\vec{u} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} }$ } \\ \text{If $u_1 + u_2 = 3$, what is $\vec{u}$? } \\ \text{Let $\vec{AB} = \dbinom{2}{5}$ } \\ \text{Let $\vec{AP} = \dbinom{x_p}{y_p}$ } \\ \text{Let line $\overline{AP}:\quad y = \dfrac{1}{2}x$ } \)

 

 

\(\mathbf{\vec{AP} = \ ? }\)

\(\begin{array}{|rcll|} \hline \overline{AB}^2 &=& 2^2 + 5^2 \\ &=& 29 \\ \overline{AP}^2 &=& x_p^2+y_p^2 \\ \overline{BP}^2 &=& (2-x_p)^2+(5-y_p)^2 \\ \overline{AP}^2 + \overline{BP}^2 &=& \overline{AB}^2 \\ x_p^2+y_p^2 + (2-x_p)^2+(5-y_p)^2 &=& 29 \\ x_p^2+y_p^2 + 4-4x_p+x_p^2+25-10y_p+y_p^2 &=& 29 \\ 2x_p^2+2y_p^2 -4x_p -10y_p+29 &=& 29 \\ 2x_p^2+2y_p^2 -4x_p -10y_p &=& 0 \quad | \quad : 2 \\ x_p^2+y_p^2 -2x_p -5y_p &=& 0 \quad | \quad y_p = \dfrac{1}{2}x_p \\ x_p^2+\dfrac{1}{4}x_p^2 -2x_p -5(\dfrac{1}{2}x_p) &=& 0 \quad | \quad : x_p \\ x_p +\dfrac{1}{4}x_p -2 -\dfrac{5}{2} &=& 0 \\ \dfrac{5}{4}x_p &=& \dfrac{9}{2} \\ \mathbf{ x_p } & \mathbf{=} & \mathbf{\dfrac{18}{5}} \\\\ y_p &=& \dfrac{1}{2}x_p \\ y_p &=& \dfrac{1}{2}\cdot\dfrac{18}{5} \\ \mathbf{ y_p } & \mathbf{=} & \mathbf{\dfrac{9}{5}} \\ \mathbf{\vec{AP}} &\mathbf{=}& \mathbf{\dfrac{9}{5}\dbinom{2}{1}} \\ \hline \end{array}\)

 

\(\text{Let $\vec{u} = \dbinom{u_1}{u_2}$ } \\ \text{Let $u^2 = u_1^2+u_2^2$ } \)

\(\begin{array}{|rcll|} \hline \dfrac{(\vec{u}\cdot \vec{AB})\cdot \vec{u}}{u^2} &=& \vec{AP} \\\\ \dfrac{\left(\dbinom{u_1}{u_2}\cdot \dbinom{2}{5}\right)\cdot \dbinom{u_1}{u_2}}{u_1^2+u_2^2} &=& \dfrac{9}{5}\dbinom{2}{1} \\\\ \dbinom{ \left( 2u_1+5u_2\right)\cdot \dfrac{u_1}{u_1^2+u_2^2} } { \left( 2u_1+5u_2\right)\cdot \dfrac{u_2}{u_1^2+u_2^2} } &=& \dfrac{9}{5}\dbinom{2}{1} \\\\ ( 2u_1+5u_2)\cdot \dfrac{u_1}{u_1^2+u_2^2} &=& \dfrac{18}{5} \\\\ 5( 2u_1+5u_2)\cdot u_1 &=& 18(u_1^2+u_2^2) \\ 5( 2u_1^2+5u_1u_2) &=& 18u_1^2+18u_2^2 \\ 10u_1^2+25u_1u_2 &=& 18u_1^2+18u_2^2 \quad | \quad u_2 = 3-u_1 \\ 10u_1^2+25u_1(3-u_1) &=& 18u_1^2+18(3-u_1)^2 \\ 10u_1^2+75u_1-25u_1^2 &=& 18u_1^2+162-108u_1+18u_1^2 \\ 51u_1^2-183u_1 + 162 &=& 0 \quad | \quad : 3 \\ 17u_1^2-61u_1 + 54 &=& 0 \\\\ u_1 &=& \dfrac{61\pm \sqrt{61^2-4\cdot 17 \cdot 54}}{2\cdot 17} \\ u_1 &=& \dfrac{61\pm \sqrt{49}}{34} \\ u_1 &=& \dfrac{61\pm 7}{34} \\\\ \hline u_1 &=& \dfrac{61+7}{34} \\ u_1 &=& \dfrac{68}{34} \\ \mathbf{u_1} & \mathbf{=} & \mathbf{2} \\\\ \quad u_2 &=& 3 - u_1 \\ \quad u_2 &=& 3 - 2 \\ \quad \mathbf{u_2} &\mathbf{=}& \mathbf{1} \quad | \quad \dfrac{u_2}{u_1} = \dfrac{1}{2} \ \checkmark \\ \hline u_1 &=& \dfrac{61-7}{34} \\ u_1 &=& \dfrac{27}{17} \\ \quad u_2 &=& 3 - u_1 \\ \quad u_2 &=& 3 - \dfrac{27}{17} \\ \quad u_2 &=& \dfrac{24}{17} \quad | \quad \dfrac{u_2}{u_1} = \dfrac{24}{27} \ne \dfrac{1}{2} \\ \hline \end{array}\)

 

\(\mathbf{\vec{u} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} }\)

 

laugh

 Feb 11, 2019
edited by heureka  Feb 12, 2019

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