How do I find the vector function of a line in 2D that goes through P(4,0) and Q(0,3)?

Question

My teacher gave us the right answer. The problem is, I can't understand it. According to my teacher, the right answer is:

$$r(t)= (1-t)\vec{P}+\vec{Q}$$

By the way, are vector equations and vector functions the same?

Answer 1

Perhaps you copied it incorrectly.

Let $\vec{P}=\langle{4,0}\rangle$, and let $\vec{Q}=\langle{0,3}\rangle$.

A corrected version of the answer you quoted would be $$r(t)=(1-t)\vec{P}+t\vec{Q}$$

An algebraically equivalent version is $$r(t)=\vec{P}+t\bigl(\vec{Q}-\vec{P}\bigr)$$ Explanation:

Subtracting any two distinct points on the line yields a nonzero vector parallel to the line, called a direction vector for the line.

So in this case, the vector $\vec{Q}-\vec{P}$ is a direction vector for the line.

Every point on the line can be obtained, starting with an initial vector which terminates on the line, say $\vec{P}$, and then adding an appropriate scalar multiple of the direction vector.

Thus, in the definition $$r(t)=\vec{P}+t\bigl(\vec{Q}-\vec{P}\bigr)$$ the variable $t$ is being used a scalar multiplier.

As regards vector functions and vector equations . . .

In this context, a vector function is a function whose output is a vector. Thus, the function $r$ is a vector function, since the output of $r$ is a vector.

A vector equation is an equation where both sides of the equation are vectors.