Calculate distance

[JJ007]

Calculate the distance from the point (-1,2,3) to the line x=2-t, y=1, z=1+t

So U=<-1,2,3> and P= (2,1,1)
PU= <-3,-1,-2>
Cross product PU x u= <1,-11,-7> ?
$\displaystyle D=\frac{||PU\ x\ u||}{||u||}=\frac {\sqrt{171}}{\sqrt{14}}?$

 

[galactus]

From the direction numbers, we know that the direction vector is $\displaystyle u=[-1,1,1]$

To find a point on a line, let t=0 and get:

$\displaystyle P=[2,1,1]$

$\displaystyle PQ=[-1-2, 2-1, 3-1]=[-3,1,2]$

Cross Product:

$\displaystyle PQ\times u=\begin{vmatrix}i&j&k\\-3&1&2\\-1&1&1\end{vmatrix}$

$\displaystyle =-i+j-2k$

$\displaystyle ||PQ\times u||=\sqrt{(-1)^{2}+1^{2}+(-2)^{2}}=\sqrt{6}$

$\displaystyle ||u||=\sqrt{3}$

$\displaystyle \frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}$

Check my figures. The method is sound, but it is easy to make an arithmetic error.

 

[JJ007]

From the direction numbers, we know that the direction vector is $\displaystyle u=[-1,1,1]$

To find a point on a line, let t=0 and get:

$\displaystyle P=[2,1,1]$

$\displaystyle PQ=[-1-2, 2-1, 3-1]=[-3,1,2]$

Cross Product:

$\displaystyle PQ\times u=\begin{vmatrix}i&j&k\\-3&1&2\\-1&1&1\end{vmatrix}$

$\displaystyle =-i+j-2k$

$\displaystyle ||PQ\times u||=\sqrt{(-1)^{2}+1^{2}+(-2)^{2}}=\sqrt{6}$

$\displaystyle ||u||=\sqrt{3}$

$\displaystyle \frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}$

Check my figures. The method is sound, but it is easy to make an arithmetic error.

For PQ I got [-3,0,0]

$\displaystyle D=\frac{\sqrt{18}}{\sqrt{3}}$

The multiple choice answers are:
$\displaystyle a.\frac{5}{\sqrt{2}}$
\(\displaystyle b.\sqrt{\frac{3}{14}\)
$\displaystyle c.\frac{5}{\sqrt{14}}$
$\displaystyle d.\sqrt{\frac{3}{2}}$
e.None of the above

I'm guessing e. ?

 

[BigGlenntheHeavy]

$\displaystyle Q \ = \ (-1,2,3), \ Line \ = \ x \ = \ 2-t,y \ = \ 1, \ z \ = \ 1+t$

$\displaystyle u \ = \ <-1,0,1>, \ when \ t \ = \ 0, \ P \ = \ (2,1,1)$

$\displaystyle PQ \ = \ <3,-1,-2>$

$\displaystyle PQ \ X \ u \ = \ <-1,-1,-1>$

$\displaystyle Hence, \ D \ = \ \frac{\sqrt3}{\sqrt2} \ = \ d$

 

[galactus]

Yep, there it is. Thanks Glenn. All it takes is a sign.