Find the equation of the plane passing through the intersection of the planes $\vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) = 7, \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and the point (2, 1, 3).

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Solution

$The equation of the plane passing through the line of intersection of the given planes is \vec{r} . (2 \hat{i} + \hat{j} + 3 \hat{k}) - 7 + λ (\vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9) = 0 \vec{r} . [(2 + 2 λ) \hat{i} + (1 + 5 λ) \hat{j} + (3 + 3 λ) \hat{k}] - 7 - 9 λ = 0 . . . (1) This passes through 2 \hat{i} + \hat{j} + 3 \hat{k} . So, (2 \hat{i} + \hat{j} + 3 \hat{k}) [(2 + 2 λ) \hat{i} + (1 + 5 λ) \hat{j} + (3 + 3 λ) \hat{k}] - 7 - 9 λ = 0 \Rightarrow 4 + 4 λ + 1 + 5 λ + 9 + 9 λ - 7 - 9 λ = 0 \Rightarrow 9 λ + 7 = 0 \Rightarrow λ = \frac{- 7}{9} Substituting this in (1), we get \vec{r} . [(2 + 2 (\frac{- 7}{9})) \hat{i} + (1 + 5 (\frac{- 7}{9})) \hat{j} + (3 + 3 (\frac{- 7}{9})) \hat{k}] - 7 - 9 (\frac{- 7}{9}) = 0 \Rightarrow \vec{r} . (4 \hat{i} - 26 \hat{j} + 6 \hat{k}) = 0 \Rightarrow \vec{r} . (2 \hat{i} - 13 \hat{j} + 3 \hat{k}) = 0$

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Find the equation of the plane passing through the intersection of the planes r→·2i^+j^+3k^=7, r→·2i^+5j^+3k^=9 and the point (2, 1, 3).

Find the equation of the plane passing through the intersection of the planes $\vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) = 7, \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and the point (2, 1, 3).