3try to find a.b and axb and b and add these three.
a.b/k=r.a and take the cross product k(rxa)+ax(rxa)=axb
prove that soln of eqn kr+rxa=b where k is non zero scaalr and a and b are non collinear vectors..is of form r=1(k2+a2)(a.bka+kb+axb)
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66Take the cross product with \vec{a} to obtain
k(\vec{a}\times\vec{r})+\vec{a}\times(\vec{r}\times\vec{a})=\vec{a}\times\vec{b}
Since \vec{a}\times(\vec{r}\times\vec{a})=a^2\vec{r}-(\vec{a}\cdot\vec{r})\vec{a}
So the above equation becomes
k(\vec{a}\times\vec{r})+a^2\vec{r}-(\vec{a}\cdot\vec{r})\vec{a}=\vec{a}\times\vec{b}
which gives us
k(\vec{r}\times\vec{a})=a^2\vec{r}-(\vec{a}\cdot\vec{r})\vec{a}-\vec{a}\times\vec{b} ------ (1)
Again the dot product with \vec{a} of the original equation, we get
k(\vec{a}\cdot \vec{r})=\vec{a}\cdot \vec{b}
So from (1) we get
\vec{r}\times \vec{a}=\dfrac{a^2}{k}\vec{r}-\dfrac{(\vec{a}\cdot\vec{b})}{k^2}\vec{a}-\dfrac{1}{k}(\vec{a}\times \vec{b})
Substitute in the original one to get the required result.