Mathematics - Vector Question with Solution | TestHub

$$\begin{array}{rl}& \mathbf{r}\times \mathbf{b}=\mathbf{c}\times \mathbf{b}\\ & \Rightarrow (\mathbf{r}-\mathbf{c})\times \mathbf{b}=0\Rightarrow \mathbf{r}-\mathbf{c}+\lambda \mathbf{b}\\ & \text{ or }\mathbf{r}=\mathbf{c}+\lambda \mathbf{b}\\ & \end{array}$$ Given, $\mathbf{r}\cdot \mathbf{a}=0$, taking dot product with a for Eq. (i). Now, $\mathbf{r}\cdot \mathbf{a}=\mathbf{a}\cdot \mathbf{c}+\lambda \mathbf{a}\cdot \mathbf{b}$ $\therefore \lambda =\frac{-\vec{a}\cdot \vec{c}}{\vec{a}\cdot \vec{b}}[\because \vec{r}\cdot \vec{a}=0]\dots$ (i) From Eqs. (i) and (ii), we get $$\mathbf{r}=\mathbf{c}-\frac{\mathbf{a}\cdot \mathbf{c}}{\mathbf{a}\cdot \mathbf{b}}\mathbf{b}$$ Taking dot with $\mathbf{b}$, we get $$\mathbf{r}\cdot \mathbf{b}=\mathbf{c}\cdot \mathbf{b}-\frac{\mathbf{a}\cdot \mathbf{c}}{\mathbf{a}\cdot \mathbf{b}}(\mathbf{b}\cdot \mathbf{b})$$ where, $\left[\begin{array}{l}\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}}\\ \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}\\ \mathbf{c}=\hat{\mathbf{i}}+2\hat{\mathbf{j}}+3\hat{\mathbf{k}}\end{array}\right]$ $$=(-1+2)-\frac{(-1-3)}{(1)}(1+1)=1+8=9$$