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Exercise 6.1 · Q1

Q.Find the vector equation of the line passing through the point having position vector −2i^+j^+k^-2\hat{i} + \hat{j} + \hat{k} and parallel to vector 4i^−j^+2k^4\hat{i} - \hat{j} + 2\hat{k}.

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✓ Free question

By Theorem 6.1, the vector equation of a line passing through a point with position vector aˉ\bar{a} and parallel to a vector bˉ\bar{b} is rˉ=aˉ+λbˉ\bar{r} = \bar{a} + \lambda\bar{b}.

Here the line passes through the point with position vector aˉ=−2i^+j^+k^\bar{a} = -2\hat{i} + \hat{j} + \hat{k}, and it is given to be parallel to bˉ=4i^−j^+2k^\bar{b} = 4\hat{i} - \hat{j} + 2\hat{k}. No further computation is needed — both pieces of data required by the formula are already given directly, so they are simply substituted into rˉ=aˉ+λbˉ\bar{r} = \bar{a} + \lambda\bar{b}.

✓Final answer

rˉ=(−2i^+j^+k^)+λ(4i^−j^+2k^)\bar{r} = (-2\hat{i} + \hat{j} + \hat{k}) + \lambda(4\hat{i} - \hat{j} + 2\hat{k})

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