A line passes through the points whose position vectors are i^+j^−2k^ and i^−3j^+k^. The position vector of a point on it at unit distance from the first point is

1 5 ( 4 i ^ + 9 j ^ − 15 k ^ )

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$A line passes through the points whose position vectors are \hat{i} + \hat{j} - 2 \hat{k} and \hat{i} - 3 \hat{j} + \hat{k} . The position vector of a$ $point on it at unit distance from the first point is$

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$\frac{1}{5} (5 \hat{i} + \hat{j} - 7 \hat{k})$

$\frac{1}{5} (4 \hat{i} + 9 \hat{j} - 15 \hat{k})$

$(\hat{i} - 4 \hat{j} + 3 \hat{k})$

$\frac{1}{5} (\hat{i} - 4 \hat{j} + 3 \hat{k})$

answer is A.

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$We have A (\hat{i} + \hat{j} - 2 \hat{k}) and B (\hat{i} - 3 \hat{j} + \hat{k})$

$On line A B points C and C^{'} are at distance 1 unit from A .$

$\vec{O C} = \vec{O A} + \vec{A C}, where \vec{A C} is unit vector in direction of \vec{A B}$

$∴ \vec{O C} = \vec{O A} + \frac{\vec{A B}}{| \vec{A B} |}$

$Similarly \vec{O C^{'}} = \vec{O A} - \frac{\vec{A B}}{| \vec{A B} |}$

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