What is vector equation of line?

Question

I know that it's given by $\vec r= \vec a + \lambda\vec b$. But why ? According to the derivation (as given in the image), I derived by observation that, the required vector equation of line is $\vec{AP}$ and it's value is $\lambda\vec b$. I derived to the previous point by the logic that, we were finding the vector equation of line that was passing through a point (P) and parallel to a vector ($\vec b$) and that vector is obviously $\vec{AP}$. But how it is said that $\vec r$ is the required vector.

Answer 1

The required vector is the position vector of the point $P$, that is the vector from the origin $O$ to $P$, i.e. : $\vec r=\vec{OP}$

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The vector equation of a line is an equation that is satisfied by the vector that has its head at a point of the line. This vector is not, in general, a vector that ''lies'' on the line, unless the line passes through the origin (that is the common starting point of all vectors).

From the figure you can see that the vector $\vec r$ is such a position vector for the point $P$. For $\lambda=0$ it coincides with the vector $\vec a$(the position vector of the point $A$) and, for any other real value of $\lambda$ its head is at a point $P$ on the line that passes through $A$ and is parallel to the vector $\vec b$.

And note that $\vec r$ is the sum ( parallelogram law) of the two vectors $\vec a$ and $\lambda \vec b$.

Answer 2

This really wants illustration but I haven't figured out a way to get sketches uploaded yet, sorry. Draw along with me as we visualize.

Imagine yourself in outer space. Space Station (Deep Space 9) is at point A. The vector from Earth (origin) to A is $a$.

A fast messenger spaceship (SpaceEx) zips by without stopping and drops off a packet of goods (the method of capturing the packet is left for a later exercise.) The path of the spaceship is a straight line for now.

The velocity of the ship is vector $b$. (good and large of course) As usual, distance travelled = velocity * time. And total distance = initial position plus distance travelled.

At time $t = 0$, the ship is just brushing past the station so its position is $a$. After one hour, its position (with respect to Earth, the origin) is $a + b$. After two hours, $a + 2b$. After three hours, $a + 3b$.

An hour before it got to the station its positon was $a - b$. Two hours before, $a - 2b$

Of course nothing restricts us to integer numbers of hours. 7.63 hours after passing the station, the ship's position with respect to Earth is $a + 7.63b$

Clearly for any real time $\lambda$ we see that the position of the ship is $a + \lambda b$.

This illustrates in concrete terms how and why the straight line equation works.

Some nice person upload a sketch?