031 Review Part 2, Problem 6

Let  ${\vec {v}}={\begin{bmatrix}-1\\3\\0\end{bmatrix}}$  and  ${\vec {y}}={\begin{bmatrix}2\\0\\5\end{bmatrix}}.$

(a) Find a unit vector in the direction of  ${\vec {v}}.$

(b) Find the distance between  ${\vec {v}}$  and  ${\vec {y}}.$

(c) Let  $L={\text{Span }}\{{\vec {v}}\}.$  Compute the orthogonal projection of  ${\vec {y}}$  onto  $L.$

Foundations:
1. The distance between the vectors  ${\vec {u}}$  and  ${\vec {v}}$  is
${\text{dist}}({\vec {u}},{\vec {v}})=||{\vec {u}}-{\vec {v}}||.$
2. The orthogonal projection of  ${\vec {y}}$  onto  $L$  is
${\hat {y}}={\text{proj}}_{L}{\vec {y}}={\frac {{\vec {y}}\cdot {\vec {u}}}{{\vec {u}}\cdot {\vec {u}}}}{\vec {u}}.$

Solution:

(a)

Step 1:
First, we calculate  $||{\vec {v}}||.$
We get
${\begin{array}{rcl}\displaystyle {||{\vec {v}}||}&=&\displaystyle {\sqrt {(-1)^{2}+3^{2}+0^{2}}}\\&&\\&=&\displaystyle {\sqrt {1+9}}\\&&\\&=&\displaystyle {{\sqrt {10}}.}\end{array}}$

Step 2:
Now, to get a unit vector in the direction of  ${\vec {v}},$  we take the vector  ${\vec {v}}$  and divide by  $||{\vec {v}}||.$
Hence, we get the vector
${\begin{array}{rcl}\displaystyle {{\frac {1}{||{\vec {v}}||}}{\vec {v}}}&=&\displaystyle {{\frac {1}{\sqrt {10}}}{\begin{bmatrix}-1\\3\\0\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}{\frac {-1}{\sqrt {10}}}\\{\frac {3}{\sqrt {10}}}\\0\end{bmatrix}}.}\end{array}}$

(b)

Step 1:

Step 2:

(c)

Step 1:

Step 2:

Final Answer:
(a)
(b)

Return to Sample Exam