# Vectors in three dimensions

The points $A$ and $B$ have position vectors, relative to the origin $O$, given by

$\vec {OA} = \vec i + \vec j +\vec k$ and $\vec {OB} = 2\vec i + 3\vec k$. The line $f$ has vector equation $\vec r = 2\vec i − 2\vec j − \vec k + \Delta(-\vec i+2\vec j+\vec k)$. The line $AB$ and line $f$ are skew. Show that the length of the perpendicular from $A$ to $f$ is $\frac1{\sqrt2}$

I know the scalar product of the direction of the perpendicular and the direction of $f$ equals zero... i have tried using this fact and the cartesian equation of $f$ to obtain the answer, my method seems plausible to me but just isnt working... can someone please show me the correct way to do this.

• Are $i,j,k$ orthonormal? What is $\Delta$? And what is the method that you're using? Apr 29 '17 at 15:01
• i, j, k are orthonormal... Δ is a variable, its just like the β in a vector equation of the form r = a + βb. I equated the cross product of the direction of the perpendicular and the direction of line f and equated to zero to find an expression in the coordinates of the point of intersection of the perpendicular with the line f, inserted the equations for the x, y and z coordinates of a general point on line f (cartesian equation) into the expression to obtain a value for Δ. Then find the actual values for the direction of the perpendicular and its magnitude. Apr 29 '17 at 20:44
• My point is that (1) your notation is not 100% standard and should be introduced (e.g. "$f = \{ \vec r = 2\vec i − 2\vec j − \vec k + \Delta(-\vec i+2\vec j+\vec k) : \Delta \in \mathbb R \}$"), more importantly (2) you can edit your question and include the clarification you gave in the comments, this way it would improve much. Apr 30 '17 at 10:16

## 1 Answer

Note: Length of perpendicular is equivalent to shortest length between two lines. It can be calculate via: $$L=(\vec a_2 - \vec a_1). \frac{(b_1 \times b_2)}{|\vec b_1 \times \vec b_2|}$$

Here, you can simply get the $$L_1=\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{2}$$

and $$L_2=\frac{x-2}{-1}=\frac{y+2}{2}=\frac{z+1}{1}$$

Therefore, $\vec a_2=2 \vec i-2\vec j-1\vec k$ and $a_1=\vec i + \vec j+ \vec k$

While you can get $b_1$ and $b_2$ via checking the direction ratios of the two lines. this should give your answer.