Respuesta :
Answer:
The answer is
[tex]\bold{ \left[\begin{array}{c}\frac{127}{27} \\ \ &\frac{65}{18} \\ \ &\frac{76}{27}\\ \ &\frac{97}{54}\\\ \end{array}\right]}[/tex]
Step-by-step explanation:
Given value:
[tex]v_1=\left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \\[/tex]
[tex]v_2=\left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] \\[/tex]
The value of [tex]v_1 \cdot v_2[/tex]:
[tex]= \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right][/tex]
The above value "[tex]v_1, v_2[/tex]" are orthogonal vectors that is: [tex]v_1 \neq 0, \ \ v_2 \neq 0[/tex]
[tex]\{v_1,v_2\}[/tex] from the above orthogonal basis of subspace w and the shortest distanc value of vector y:
[tex]=\left[\begin{array}{c}3&9&-5&8\\\ \end{array}\right] \\\\[/tex]
The value of y on [tex]W= \frac{y \cdot v_1}{v_1 \cdot v_1} \times v1 + \frac{y \cdot v_2}{v_2 \cdot v_2} \times v_2[/tex]
[tex]= \frac{\left[\begin{array}{c}3&9&-5&8\\\ \end{array}\right] \cdot \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] }{\left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \cdot \left[\begin{array}{c}5&4&3&2\\ \end{array}\right]} \times \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] +[/tex] [tex]\frac{\left[\begin{array}{c}3&9&5&-8\\\ \end{array}\right] \cdot \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] }{\left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] \cdot \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]} \times \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right][/tex]
[tex]= \frac{50}{54} \left[\begin{array}{c}5&4&3&2\\\ \end{array}\right] - \frac{1}{54} \left[\begin{array}{c}-4&5&-2&3\\\ \end{array}\right]\\\\[/tex]
[tex]= \left[\begin{array}{c}\frac{127}{27} \\ \ &\frac{65}{18} \\ \ &\frac{76}{27}\\ \ &\frac{97}{54}\\\ \end{array}\right][/tex]
