If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$, then the vector $r$ satisfying $r imes b=c imes b$ and $r . a=0$ is

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If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$, then the vector $r$ satisfying $r 	imes b=c 	imes b$ and $r . a=0$ is

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(d) Three vectors $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are given as, $a=2 \hat{i}+\hat{k}, b=\hat{i}+\hat{j}+\hat{k}$ and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$ Given condition, $\mathrm{r} \times \mathrm{b}=\mathrm{c} \times \mathrm{b}$ $\begin{aligned} & \mathrm{r} \times \mathrm{b}-\mathrm{c} \times \mathrm{b}=0 \\ & (\mathrm{r}-\mathrm{c}) \times \mathrm{b}=0 \end{aligned}$ It means, $(\mathrm{r}-\mathrm{c}) \| \mathrm{b}$ So, $r-c=\lambda b$

Also given, $\mathrm{r} \cdot \mathrm{a}=0$ $(c+\lambda b) \cdot a=0$ [using Eq. (i)] $\begin{aligned} & (4 \hat{i}-3 \hat{j}+7 \hat{k}+\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}) \cdot(2 \hat{i}+\hat{k})=0 \\ & {[(4+\lambda) \cdot \hat{i}+(-3+\lambda) \hat{j}+(7+\lambda) \hat{k}] \cdot(2 \hat{i}+\hat{k})=0} \\ & (4+\lambda) \cdot 2+(7+\lambda) \cdot 1=0 \\ & 8+2 \lambda+7+\lambda=0 \\ & 3 \lambda=-15 \\ & \therefore \quad \lambda=-5 \end{aligned}$ Put the value of $\lambda$ in Eq. (i), we get $\begin{aligned} & r=4 \hat{i}-3 \hat{j}+7 \hat{k}-5(\hat{i}+\hat{j}+\hat{k}) \\ & =4 \hat{i}-3 \hat{j}+7 \hat{k}-5 \hat{i}-5 \hat{j}-5 \hat{k}=-\hat{i}-8 \hat{j}+2 \hat{k} \end{aligned}$

If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{k}},…

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