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

Question

Accepted Answer

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> [!TLDR]
> Write the family of planes through the line of intersection of the two given planes using a parameter $\lambda$, then fix $\lambda$ by requiring the plane to pass through $(2,1,3)$.
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> **Step 1 — write the family of planes through the intersection**
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> For planes $\vec r\cdot\vec n_1 = d_1$ and $\vec r\cdot\vec n_2 = d_2$, every plane through their line of intersection has the form $\vec r\cdot(\vec n_1+\lambda\vec n_2) = d_1+\lambda d_2$. Here:
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> $$\vec r\cdot\big[(2\hat i+2\hat j-3\hat k) + \lambda(2\hat i+5\hat j+3\hat k)\big] = 7+9\lambda$$
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> $$\vec r\cdot\big[(2+2\lambda)\hat i + (2+5\lambda)\hat j + (-3+3\lambda)\hat k\big] = 7+9\lambda$$
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> **Step 2 — impose the condition that it passes through $(2,1,3)$**
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> Here $\vec r = 2\hat i+\hat j+3\hat k$. Substituting:
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> $$2(2+2\lambda) + 1(2+5\lambda) + 3(-3+3\lambda) = 7+9\lambda$$
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> $$(4+4\lambda)+(2+5\lambda)+(-9+9\lambda) = 7+9\lambda \implies -3+18\lambda = 7+9\lambda$$
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> $$9\lambda = 10 \implies \lambda = \frac{10}{9}$$
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> **Step 3 — substitute $\lambda$ back into the family**
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> $$2+2\lambda = 2+\frac{20}{9}=\frac{38}{9}, \quad 2+5\lambda = 2+\frac{50}{9}=\frac{68}{9}, \quad -3+3\lambda = -3+\frac{30}{9}=\frac13, \quad 7+9\lambda = 7+10=17$$
> …

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

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

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