Q.19 Find the vector equation of the plane passing through the point \(A(-1, 2, -5)\) and parallel to the vectors \(4\hat{i} - \hat{j} + 3\hat{k}\) and \(\hat{i} + \hat{j} - \hat{k}\).
Best answer
Answer: \(\bar{r} = (-\hat{i} + 2\hat{j} - 5\hat{k}) + \lambda(4\hat{i} - \hat{j} + 3\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})\), i.e. \(\bar{r}\cdot(-2\hat{i} + 7\hat{j} + 5\hat{k}) = -9\)
The plane through \(A(\bar{a})\) parallel to \(\bar{b}\) and \(\bar{c}\) is \(\bar{r} = \bar{a} + \lambda\bar{b} + \mu\bar{c}\), with \(\bar{a} = -\hat{i} + 2\hat{j} - 5\hat{k}\), \(\bar{b} = 4\hat{i} - \hat{j} + 3\hat{k}\), \(\bar{c} = \hat{i} + \hat{j} - \hat{k}\).
Normal form: \(\bar{n} = \bar{b} \times \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(1 - 3) - \hat{j}(-4 - 3) + \hat{k}(4 + 1) = -2\hat{i} + 7\hat{j} + 5\hat{k}\).
\(\bar{a}\cdot\bar{n} = 2 + 14 - 25 = -9\). So the equation is \(\bar{r}\cdot(-2\hat{i} + 7\hat{j} + 5\hat{k}) = -9\), i.e. \(\bar{r}\cdot(2\hat{i} - 7\hat{j} - 5\hat{k}) = 9\).
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