Finding the length of the projection of the vector onto a line using parametric equation

Question

What is the length of the projection of the vector $(3, 4,-4)$ onto a line whose parametric equation is the following?

$$\begin{aligned} x &= 2t + 1\\ y &= -t + 3\\ z &= t - 1\end{aligned}$$

Hint: find a unit vector in the direction of the line and construction its projection operators.

Answer 1

Hint. Your line is parallel to the vector $(2,-1,1)$, the unit vector along which is $\displaystyle(\frac2{\sqrt6},\frac{-1}{\sqrt6},\frac1{\sqrt6})$. The projection of vector $\vec A$ along $\vec B$ is given by $\displaystyle\frac{\vec A\cdot\vec B}{|\vec B|}$.

Answer 2

To get the length of the projection of this vector on the line yes you first get a unit directing vector to the line.

$\vec v(2,-1,1)$ is a directing vector the line its magnitude is $||\vec v||=\sqrt 6$

So a unit directing vector to the line will be $\vec {v'} (\frac{2}{\sqrt 6},\frac{-1}{\sqrt 6},\frac{1}{\sqrt 6})$

Now just do the dot product $|(3,4,-4).\vec {v'}|=|\frac{6}{\sqrt 6} -\frac{4}{\sqrt 6} -\frac{4}{\sqrt 6}|=\frac{2}{\sqrt 6}$

And thats the length of the projection.

Now why does that work

In general, $\vec u.\vec v= u.v.\cos \theta$ (where $\theta$ is the angle between the 2 vectors)

Draw 2 vectors $\vec u$ and $\vec v$ and draw the projection of $\vec v$ on $\vec u$ for example.

You'll have a right triangle, and you'll find that $\cos \theta=\frac{projection}{v}$ (I mean by 'projection' the length of the projection of $\vec v$ on $\vec u$)

Substitute above and you'll get that $\vec u.\vec v=u.{projection}$

But in your case you have a unit vector so $u=1$

Thats why $\vec u.\vec v={projection}$

Answer 3

One parameterisation of the line is given by $$\mathbf{x}(t) = t\begin{pmatrix} 2 \\ -1 \\ 1\end{pmatrix} + \begin{pmatrix}1 \\ 3 \\ 1 \end{pmatrix}, \ t \in \mathbb{R}.$$ The geometrical interpretation of $\mathbf{x}$ is that for every real number $t$, $\mathbf{x}(t)$ corresponds to an arrow which points to a point on the line with its root at the origin in $\mathbb{R}^3$.

In fact, the outputs of the function $\mathbf{x}$ collectively point to every point on the line.

In order to find the length of the projection of $(3,4,-4)$ onto the line, we first find a unit vector which points along the line, and then take the magnitude of the component of $(3,4,-4)$ in the direction of that unit vector. To find a unit vector which points along the line, we find two different vectors which point to different points on the line, take their difference, and then normalise the result.

Finding two different vectors which point to the line is the same as finding two different outputs of $\mathbf{x}$. For example, set $t = 0$, to obtain $$\mathbf{x}(0) = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}.$$ Then we can for instance set $t = 1$, to obtain $$\mathbf{x}(1) = \begin{pmatrix} 2 \\-1 \\\ 1\end{pmatrix}+ \begin{pmatrix}1 \\ 3 \\\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}.$$

Now, the vector pointing from $\mathbf{x}(0)$ to $\mathbf{x}(1)$ which we will call $\mathbf{n}$ is given by $$\mathbf{n} = \mathbf{x}(1) - \mathbf{x}(0) = \begin{pmatrix} 2 \\ -1 \\ 1\end{pmatrix}.$$ We then obtain the unit vector pointing along $\mathbf{n}$ which we call $\hat{\mathbf{n}}$ by setting $$\hat{\mathbf{n}} = \frac{\mathbf{n}}{||\mathbf{n}||} = \frac{1}{\sqrt{6}}\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}.$$ The magnitude of the component of $(3,4,-4)$ in the direction of the line, i.e. the length of the projection of $(3,4,-4)$ onto the line, is given by $|(3,4,-4) \cdot \hat{\mathbf{n}}| = \frac{\sqrt{6}}{3}.$

I suggest checking out this series of videos to help you nail the intuitions behind vectors, dot products and much more.