1
$\begingroup$

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 $\vec{r}=\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|}$.

Please help me.Thanks.

$\endgroup$
3
$\begingroup$

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 $\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.