Equation of Straight Line in Space/Vector Form/Also presented as

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Vector Equation of Straight Line in Space: Also presented as

Let $a$ and $b$ be points in cartesian $3$-space with position vectors $\mathbf a$ and $\mathbf b$ respectively.


The vector form of the equation of a straight line in space can be presented in the forms:

$\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$

or:

$\paren {\mathbf r - \mathbf a} \times \paren {\mathbf b - \mathbf a} = \mathbf 0$

which is the same straight line.


Proof

We have that $\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$ is exactly the same form as $\mathbf r = \mathbf a + t \mathbf b$ but substituting $\mathbf b - \mathbf a$ for $\mathbf b$.

From the exposition of the proof, $\mathbf r = \mathbf a + t \mathbf b$ is all the points in the straight line passing through $a$ parallel to $\mathbf b$.

Hence $\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$ is all the points in the straight line passing through $a$ parallel to $\mathbf b - \mathbf a$.

But $\mathbf b - \mathbf a$ is the vector that takes you:

from $a$, whose position vector is $\mathbf a$
to $b$, whose position vector is $\mathbf b$.

Hence $b$ is itself on the straight line passing through $\mathbf a$ parallel to $\mathbf b - \mathbf a$.

Hence $\mathbf r = \mathbf a + t \paren {\mathbf b - \mathbf a}$ is the straight line passing through $a$ and $b$.

$\Box$


We note that, by Cross Product of Parallel Vectors:

$\paren {\mathbf r - \mathbf a} \times \paren {\mathbf b - \mathbf a} = \mathbf 0$

defines the set of points in $\R^3$ passing through $\mathbf a$ parallel to $\mathbf b - \mathbf a$.

That is, from the above, the set of points in the straight line passing through $a$ and $b$.

$\blacksquare$


Sources

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