Let hat{i}-hat{j}+2 hat{k} and hat{i}+2 hat{j | vedclass

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(A) Given position vectors are $\overrightarrow{OA} = \hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{OB} = \hat{i}+2\hat{j}-2\hat{k}$.Any point $C$ on

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$\hat{i}+8 \hat{j}-10 \hat{k}$

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(A) Given position vectors are $\overrightarrow{OA} = \hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{OB} = \hat{i}+2\hat{j}-2\hat{k}$.Any point $C$ on the line passing through $A$ and $B$ can be represented as $\overrightarrow{OC} = \overrightarrow{OA} + t(\overrightarrow{OB} - \overrightarrow{OA})$.$\overrightarrow{OB} - \overrightarrow{OA} = (\hat{i}+2\hat{j}-2\hat{k}) - (\hat{i}-\hat{j}+2\hat{k}) = 3\hat{j} - 4\hat{k}$.So,$\overrightarrow{OC} = (\hat{i}-\hat{j}+2\hat{k}) + t(3\hat{j}-4\hat{k}) = \hat{i} + (3t-1)\hat{j} + (2-4t)\hat{k}$.Given $BC = 10$,we have $|\overrightarrow{OC} - \overrightarrow{OB}| = 10$.$\overrightarrow{OC} - \overrightarrow{OB} = (\hat{i} + (3t-1)\hat{j} + (2-4t)\hat{k}) - (\hat{i}+2\hat{j}-2\hat{k}) = (3t-3)\hat{j} + (4-4t)\hat{k}$.$|\overrightarrow{OC} - \overrightarrow{OB}|^2 = (3t-3)^2 + (4-4t)^2 = 10^2 = 100$.$9(t-1)^2 + 16(1-t)^2 = 100 \Rightarrow 25(t-1)^2 = 100 \Rightarrow (t-1)^2 = 4$.$t-1 = \pm 2$,so $t = 3$ or $t = -1$.For $t=3$,$\overrightarrow{OC} = \hat{i} + (3(3)-1)\hat{j} + (2-4(3))\hat{k} = \hat{i} + 8\hat{j} - 10\hat{k}$.

Let $\hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-2 \hat{k}$ be the position vectors of points $A$ and $B$ respectively. If $C$ is a point on the line joining $A$ and $B$ such that $BC=10$,then the position vector of $C$ can be

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