Find the point on the line (0,0) + a*(3, 1) that is nearest to the point (1, 4).

Step-by-step solution

1. 1

   Use orthogonal projection. The nearest point on a line through the origin is the projection of $\mathbf p$ onto the direction $\mathbf d$:

   $\operatorname{proj}_{\mathbf d}\mathbf p=\frac{\mathbf p\cdot\mathbf d}{\mathbf d\cdot\mathbf d}\,\mathbf d$

   The characterising property is that the residual $\mathbf p-\operatorname{proj}_{\mathbf d}\mathbf p$ is perpendicular to $\mathbf d$; that perpendicularity is what makes the distance smallest.

2. 2

   Compute the two dot products. With $\mathbf p=(1,4)$ and $\mathbf d=(3,1)$:

   $\mathbf p\cdot\mathbf d=1(3)+4(1)=7,\qquad \mathbf d\cdot\mathbf d=9+1=10$

3. 3

   Find the parameter and the point.

   $a=\frac{7}{10}=0.7\;\Longrightarrow\;\text{nearest point}=0.7(3,1)=\left(\frac{21}{10},\,\frac{7}{10}\right)=(2.1,\,0.7)$

4. 4

   Verify the perpendicularity. The residual vector from the foot to $P$ is

   $(1,4)-(2.1,0.7)=(-1.1,\,3.3)$

   and its dot product with the direction is $(-1.1)(3)+(3.3)(1)=-3.3+3.3=0\;\checkmark$. A zero dot product is proof, not evidence — no other point on the line has this property.

5. 5

   Read off the minimum distance as a by-product.

   $\left\|(-1.1,\,3.3)\right\|=\sqrt{1.21+10.89}=\sqrt{12.1}\approx 3.4785$

   Testing neighbouring parameters confirms it is a minimum: at $a=0.6$ the distance is $\sqrt{12.2}\approx 3.493$ and at $a=0.8$ it is also $\sqrt{12.2}$, both larger than $\sqrt{12.1}$.