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Let $\hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-2 \hat{k}$ be the position vectors of points $\mathrm{A}$ and $\mathrm{B}$ respectively. If $\mathrm{C}$ is a point on the line joining $\mathrm{A}$ and $\mathrm{B}$ such that $\mathrm{BC}=10$, then the position vector of $\mathrm{C}$ can be
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The correct answer is:
$\hat{\mathrm{i}}+8 \hat{\mathrm{j}}-10 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
$\begin{aligned} & \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{OA}}+t(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}})=(\hat{\mathrm{i}}-\hat{j}+2 \hat{\mathrm{k}})+\mathrm{t}(3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{OC}}=\hat{\mathrm{i}}+(3 \mathrm{t}-1) \hat{\mathrm{j}}+(2-4 \mathrm{t}) \hat{\mathrm{k}} \\ & \because \quad|\overrightarrow{\mathrm{BC}}|=10 \\ & \Rightarrow|\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}|=10\end{aligned}$
$\begin{aligned} & \Rightarrow|0 \hat{i}+(3 t-1-2) \hat{j}+(2-4 t+2) \hat{k}|=10 \\ & \Rightarrow(3 t-3)^2+(4-4 t)^2=100 \\ & \Rightarrow 9(t-1)^2+16(t-1)^2=100 \\ & \Rightarrow 25(t-1)^2=100 \Rightarrow(t-1)^2=4 \\ & \Rightarrow t-1=2 \Rightarrow t=3 \\ & \therefore \overrightarrow{O C}=\hat{i}+8 \hat{j}-10 \hat{k}\end{aligned}$
$\begin{aligned} & \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{OA}}+t(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}})=(\hat{\mathrm{i}}-\hat{j}+2 \hat{\mathrm{k}})+\mathrm{t}(3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{OC}}=\hat{\mathrm{i}}+(3 \mathrm{t}-1) \hat{\mathrm{j}}+(2-4 \mathrm{t}) \hat{\mathrm{k}} \\ & \because \quad|\overrightarrow{\mathrm{BC}}|=10 \\ & \Rightarrow|\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}|=10\end{aligned}$
$\begin{aligned} & \Rightarrow|0 \hat{i}+(3 t-1-2) \hat{j}+(2-4 t+2) \hat{k}|=10 \\ & \Rightarrow(3 t-3)^2+(4-4 t)^2=100 \\ & \Rightarrow 9(t-1)^2+16(t-1)^2=100 \\ & \Rightarrow 25(t-1)^2=100 \Rightarrow(t-1)^2=4 \\ & \Rightarrow t-1=2 \Rightarrow t=3 \\ & \therefore \overrightarrow{O C}=\hat{i}+8 \hat{j}-10 \hat{k}\end{aligned}$
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