Find the distance between the point (1, 4) and the line (0,0) + a*(3, 1).

Step-by-step solution

1. 1

   Write the distance to a general point of the line. A point of $\ell$ is $(3a,a)$, so

   $d(a)=\sqrt{(1-3a)^2+(4-a)^2}=\sqrt{1-6a+9a^2+16-8a+a^2}=\sqrt{10a^2-14a+17}$

   This function answers "how far is $P$ from the point at parameter $a$" — the distance to the line is its minimum value.

2. 2

   Minimise the quadratic under the root. A square root is increasing, so minimising $d(a)$ is the same as minimising $q(a)=10a^2-14a+17$, whose vertex is at

   $a=-\frac{-14}{2(10)}=\frac{14}{20}=0.7$

   This is the same $a$ the perpendicular projection would give: $\tfrac{\mathbf p\cdot\mathbf d}{\mathbf d\cdot\mathbf d}=\tfrac{3+4}{10}=0.7$.

3. 3

   Evaluate at the vertex.

   $q(0.7)=10(0.49)-14(0.7)+17=4.9-9.8+17=12.1\;\Longrightarrow\;d=\sqrt{12.1}\approx 3.4785$

4. 4

   Confirm with the 2D cross-product shortcut. In the plane the distance from $P$ to a line through the origin with direction $\mathbf d=(3,1)$ is

   $d=\frac{\left|p_xd_y-p_yd_x\right|}{\|\mathbf d\|}=\frac{|1\cdot 1-4\cdot 3|}{\sqrt{10}}=\frac{11}{\sqrt{10}}\approx 3.4785$

   and indeed $\left(\tfrac{11}{\sqrt{10}}\right)^2=\tfrac{121}{10}=12.1\;\checkmark$, matching the minimised quadratic exactly.

5. 5

   State the answer carefully. If the question means the distance to the whole line, the answer is

   $d=\frac{11}{\sqrt{10}}=\frac{11\sqrt{10}}{10}\approx 3.4785$

   If instead a specific parameter $a$ is intended, the distance is the function $\sqrt{10a^2-14a+17}$, which equals $\sqrt{17}\approx 4.12$ at $a=0$ and grows without bound as $|a|\to\infty$.