We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
157
1
avatar

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
avatar+22547 
+5

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

12 Online Users

avatar
avatar