simple vector problem Let V=(0,3,2) and A=(-1,0,1). Let P be the point on the line passing through A with direction vector (1,1,1) that is closest to V. Find AP, P, and the distance between V and the line.
This is a pretty basic question, but I haven't done any problem like these. Can someone give me a pointer?
For my try, I got an answer of (3,0,2) which is wrong.
$VA = (-1,0,1)-(0,3,2) = (-1,-3,-1)$
$VA\cdot \frac {(1,1,1)}{\|(1,1,1)\|}$ will give the magnitude of the orthogonal projection of V onto the line.
$PA = VA\cdot \frac {(1,1,1)}{\|(1,1,1)\|^2}(1,1,1)$
$PA = (-\frac {5}{3},-\frac {5}{3},-\frac {5}{3})$
$PA$ and $VA$ form one leg and the hypotenuse of a right triangle.
$\|PA\|^2 + \|PV\|^2 = \|VA\|^2$
$\frac {25}{3} + \|PV\|^2 = 11\\
\|PV\|^2 = \frac {8}{3}
\|PV\| = \sqrt {\frac {8}{3}}$
The equation of line containing P is
$\textbf r=(-1,0,1)+\lambda (1,1,1) $
This gives the point P as $(\lambda-1,\lambda,\lambda+1)$
Now P is nearer to V when the direction ratio of vector joining V and P is perpendicular to direction ratio of the line.
This gives,
$(\lambda-1,\lambda-3,\lambda-1)\cdot(1,1,1)=0$
Using this you can find $\lambda$ and hence the point P.