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Exercise 6.1 · Q2

Q.Find the vector equation of the line passing through points having position vectors 3i^+4j^−7k^3\hat{i} + 4\hat{j} - 7\hat{k} and 6i^−j^+k^6\hat{i} - \hat{j} + \hat{k}.

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By Theorem 6.3, the vector equation of a line passing through two points A(aˉ)A(\bar{a}) and B(bˉ)B(\bar{b}) is rˉ=aˉ+λ(bˉ−aˉ)\bar{r} = \bar{a} + \lambda(\bar{b} - \bar{a}).

Here aˉ=3i^+4j^−7k^\bar{a} = 3\hat{i} + 4\hat{j} - 7\hat{k} and bˉ=6i^−j^+k^\bar{b} = 6\hat{i} - \hat{j} + \hat{k}. First compute the direction bˉ−aˉ\bar{b} - \bar{a}:

bˉ−aˉ=(6i^−j^+k^)−(3i^+4j^−7k^)=(6−3)i^+(−1−4)j^+(1+7)k^=3i^−5j^+8k^.\bar{b} - \bar{a} = (6\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} + 4\hat{j} - 7\hat{k}) = (6-3)\hat{i} + (-1-4)\hat{j} + (1+7)\hat{k} = 3\hat{i} - 5\hat{j} + 8\hat{k}.

Substituting aˉ\bar{a} and this direction into the formula gives the required equation. As a check, putting λ=1\lambda = 1 should reproduce bˉ\bar{b}: (3i^+4j^−7k^)+(3i^−5j^+8k^)=6i^−j^+k^(3\hat{i}+4\hat{j}-7\hat{k}) + (3\hat{i}-5\hat{j}+8\hat{k}) = 6\hat{i}-\hat{j}+\hat{k}, which indeed matches bˉ\bar{b}.

✓Final answer

rˉ=(3i^+4j^−7k^)+λ(3i^−5j^+8k^)\bar{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda(3\hat{i} - 5\hat{j} + 8\hat{k})

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