Shortest distance of Point $A$ from $2\vec{OA} - \vec{OB}$ The position vectors of points $A$ and $B$ are $3i-k$and $2i + j + 3k$ respectively. Find the shortest distance of Point $A$ from $2\vec{OA} - \vec{OB}$ 
I found the vector and subsequently the magnitude of vector $2\vec{OA} - \vec{OB}= \sqrt {42}$ units 
i drew this to help me understand the question -

I know the magnitude of vector 2OA-OB 
now, I am left to find the length of projection of $2\vec{OA} - \vec{OB}$ onto $A = \frac{17}{\sqrt {10}}$.
I then used Pythagoras theorem to find the shortest distance - 
$ \sqrt { (\sqrt42)^2 - (\frac{17}{\sqrt{10}})^2 } $.
However, my answer is wrong. Have I gotten anything wrong here ? 
 A: Your method is fine but you have projected $(2OA-OB)$ ont $OA$ insted of $OA$ onto $2(OA-OB)$.
In this way we obtain


*

*$A=(3,0,-1)$

*$B=(2,1,3)$

*$2OA-OB=(4,-1,-5)$


then the projection $|OP|$ of $OA$ onto $2OA-OB$ is
$$|OP|=OA\cdot \frac{2OA-OB}{|2OA-OB|}=\frac{17}{\sqrt{42}}$$
then
$$|D|= (|OA|^2-|OP|^2)^\frac12=\left(10-\frac{289}{42}\right)^\frac12=\left(\frac{131}{42}\right)^\frac12=$$
A: I want to offer an alternative. We agree that 2OA - OB is $c=(4,-1,5)$  We are asked to find the vector projection of OA onto $c=(4,-1,5)$. 
$$P=\mathbf{c\cdot}\frac{\mathbf{c\cdot\overline{OA}}}{\Vert\mathbf{c}\Vert^{2}}=\frac{\left(\begin{array}{c}4\\-1\\5\end{array}\right)\cdot\left[\left(\begin{array}{c}4\\-1\\5
\end{array}\right)\cdot\left(\begin{array}{c}3\\0\\1\end{array}\right)\right]}{\left\Vert \left(\begin{array}{c}4\\-1\\5
\end{array}\right)\right\Vert ^{2}}=\frac{\left(\begin{array}{c}4\\-1\\5\end{array}\right)\cdot 17}{42}=\left(\begin{array}{c}34/21\\-17/42\\-85/42\end{array}\right)$$
From here we are asked to find the distance from point A to point P.
$$dist=\sqrt{\left(\frac{34}{21}-3\right)^{2}+\left(\frac{-17}{42}-0\right)^{2}+\left(\frac{-85}{42}--1\right)^{2}}\\=\sqrt{\left(\frac{29}{21}\right)^{2}+\left(\frac{-17}{42}\right)^{2}+\left(\frac{-43}{42}\right)^{2}}=\sqrt{\frac{131}{42}}$$  
