Calculate distance
- Thread starter JJ007
- Start date
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.
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.
galactus said:
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
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- Mar 8, 2009
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\(\displaystyle Q \ = \ (-1,2,3), \ Line \ = \ x \ = \ 2-t,y \ = \ 1, \ z \ = \ 1+t\)
\(\displaystyle u \ = \ <-1,0,1>, \ when \ t \ = \ 0, \ P \ = \ (2,1,1)\)
\(\displaystyle PQ \ = \ <3,-1,-2>\)
\(\displaystyle PQ \ X \ u \ = \ <-1,-1,-1>\)
\(\displaystyle Hence, \ D \ = \ \frac{\sqrt3}{\sqrt2} \ = \ d\)
\(\displaystyle u \ = \ <-1,0,1>, \ when \ t \ = \ 0, \ P \ = \ (2,1,1)\)
\(\displaystyle PQ \ = \ <3,-1,-2>\)
\(\displaystyle PQ \ X \ u \ = \ <-1,-1,-1>\)
\(\displaystyle Hence, \ D \ = \ \frac{\sqrt3}{\sqrt2} \ = \ d\)
