Exercise 6.2.20

If $v$ is the orthogonal projection of $u$, then the distance of $u$ is the length of $u-v$.

1.

The distance is

$$\parallel (2,6)-\frac{26}{17}(1,4)\parallel =\frac{2}{\sqrt{17}}.$$

2.

The distance is

$$\parallel (2,1,3)-\frac{1}{14}(29,17,40)\parallel =\frac{1}{\sqrt{14}}.$$

3.

The distance is

$$\parallel (-2x^2+3x+4)-\frac{1}{9}(x+1)\parallel =\sqrt{\frac{30527}{1215}}.$$