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Question: A line passes through the points whose position vectors are $\hat{i} + \hat{j} - 2\hat{k}$ and $\hat...

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

A

i^+15j^−75k^\hat{i} + \frac{1}{5}\hat{j} - \frac{7}{5}\hat{k}

B

i^+95j^−135k^\hat{i} + \frac{9}{5}\hat{j} - \frac{13}{5}\hat{k}

C

i^+j^−2k^\hat{i} + \hat{j} - 2\hat{k}

D

i^−3j^+k^\hat{i} - 3\hat{j} + \hat{k}

Answer

i^+15j^−75k^\hat{i} + \frac{1}{5}\hat{j} - \frac{7}{5}\hat{k} or i^+95j^−135k^\hat{i} + \frac{9}{5}\hat{j} - \frac{13}{5}\hat{k}

Explanation

Solution

Let the position vectors be a⃗=i^+j^−2k^\vec{a} = \hat{i} + \hat{j} - 2\hat{k} and b⃗=i^−3j^+k^\vec{b} = \hat{i} - 3\hat{j} + \hat{k}. The direction vector is d⃗=b⃗−a⃗=−4j^+3k^\vec{d} = \vec{b} - \vec{a} = -4\hat{j} + 3\hat{k}. The unit direction vector is d^=−4j^+3k^5\hat{d} = \frac{-4\hat{j} + 3\hat{k}}{5}. The position vector of a point at unit distance from a⃗\vec{a} is r⃗=a⃗±1⋅d^\vec{r} = \vec{a} \pm 1 \cdot \hat{d}. So, r⃗1=(i^+j^−2k^)+−4j^+3k^5=i^+15j^−75k^\vec{r}_1 = (\hat{i} + \hat{j} - 2\hat{k}) + \frac{-4\hat{j} + 3\hat{k}}{5} = \hat{i} + \frac{1}{5}\hat{j} - \frac{7}{5}\hat{k}. And, r⃗2=(i^+j^−2k^)−−4j^+3k^5=i^+95j^−135k^\vec{r}_2 = (\hat{i} + \hat{j} - 2\hat{k}) - \frac{-4\hat{j} + 3\hat{k}}{5} = \hat{i} + \frac{9}{5}\hat{j} - \frac{13}{5}\hat{k}.