Question

Find the closest point to y in the subspace W spanned by v1 and v

Answer 1

Answer:

The answer is

$\bold{ \left[\begin{array}{c}\frac{127}{27} \\ \ &\frac{65}{18} \\ \ &\frac{76}{27}\\ \ &\frac{97}{54}\\\ \end{array}\right]}$

Step-by-step explanation:

Given value:

$v_1=\left[\begin{array}{c}5&4&3&2\\ \end{array}\right]$

$v_2=\left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]$    

The value of $v_1 \cdot v_2$:

$= \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]$

The above value "$v_1, v_2$" are orthogonal vectors that is: $v_1 \neq 0, \ \ v_2 \neq 0$

$\{v_1,v_2\}$ from the above orthogonal basis of subspace w and the shortest distanc value of vector y:

$=\left[\begin{array}{c}3&9&-5&8\\\ \end{array}\right]$

The value of y on $W= \frac{y \cdot v_1}{v_1 \cdot v_1} \times v1 + \frac{y \cdot v_2}{v_2 \cdot v_2} \times v_2$

$= \frac{\left[\begin{array}{c}3&9&-5&8\\\ \end{array}\right] \cdot \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] }{\left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \cdot \left[\begin{array}{c}5&4&3&2\\ \end{array}\right]} \times \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] +$ $\frac{\left[\begin{array}{c}3&9&5&-8\\\ \end{array}\right] \cdot \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] }{\left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] \cdot \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]} \times \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]$

$= \frac{50}{54} \left[\begin{array}{c}5&4&3&2\\\ \end{array}\right] - \frac{1}{54} \left[\begin{array}{c}-4&5&-2&3\\\ \end{array}\right]$

$= \left[\begin{array}{c}\frac{127}{27} \\ \ &\frac{65}{18} \\ \ &\frac{76}{27}\\ \ &\frac{97}{54}\\\ \end{array}\right]$