# Find the vector which is equally inclined to $\vec{a},\vec{b},\vec{c}$ vectors.

If $\vec{a},\vec{b},\vec{c}$ are three mutually perpendicular vectors,then find the vector which is equally inclined to $\vec{a},\vec{b},\vec{c}$ vectors.

My Attempt:
Since $\vec{a},\vec{b},\vec{c}$ are three mutually perpendicular vectors,so they are linearly independent vectors.So we can express desired vector $\vec{r}$ in terms of $\vec{a},\vec{b},\vec{c}$.
$\vec{r}=x\vec{a}+y\vec{b}+z\vec{c}$
We need to find $x,y,z$.
Let the angle between $\vec{r}$ and $\vec{a},$between $\vec{r}$ and $\vec{b}$,between $\vec{r}$ and $\vec{c}$ is $\theta.$
$\vec{r}.\vec{a}=x|\vec{a}|^2=|\vec{r}||\vec{a}|\cos\theta$
$x=\frac{|\vec{r}|\cos\theta}{|\vec{a}|}$
Similarly,$y=\frac{|\vec{r}|\cos\theta}{|\vec{b}|}$
$z=\frac{|\vec{r}|\cos\theta}{|\vec{c}|}$
So $\vec{r}=\frac{|\vec{r}|\cos\theta}{|\vec{a}|}\vec{a}+\frac{|\vec{r}|\cos\theta}{|\vec{b}|}\vec{b}+\frac{|\vec{r}|\cos\theta}{|\vec{c}|}\vec{c}$
But i cannot get rid of $|\vec{r}|\cos\theta$ from the expression.Answer should be in terms of $\vec{a},\vec{b},\vec{c}$ only.Answer given is $\vec{r}=\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|}$.

Your answer is almost correct, you just need to factor out $|\vec{r}|\cos\theta$.
$\vec{r}=\frac{|\vec{r}|\cos\theta}{|\vec{a}|}\vec{a}+\frac{|\vec{r}|\cos\theta}{|\vec{b}|}\vec{b}+\frac{|\vec{r}|\cos\theta}{|\vec{c}|}\vec{c}= |\vec{r}|\cos\theta(\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|})$.
Note that the question does not ask for any specific length so the scalar multiplier can be ignored giving the answer $\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|}$.