Let π₁ and π₂ be two planes with equations \(\vec{r}.{{\hat{n}}_{1}}={{d}_{1}}\) and \(\vec{r}.{{\hat{n}}_{2}}={{d}_{2}}\), respectively. The position vector of any point on the line of intersection must satisfy both the equation.  If \(\vec{t}\) is the position vector of a point on the line, then \(\vec{t}.{{\hat{n}}_{1}}={{d}_{1}}\) and \(\vec{t}.{{\hat{n}}_{2}}={{d}_{2}}\). Therefore, for all real values of λ, we have \(\vec{t}.\left( {{{\hat{n}}}_{1}}+\lambda {{{\hat{n}}}_{2}} \right)={{d}_{1}}+\lambda {{d}_{2}}\)… (1) Since \(\vec{t}\) is arbitrary, it satisfies for any point on the line. Hence, the equation \(\vec{r}.\left( {{{\hat{n}}}_{1}}+\lambda {{{\hat{n}}}_{2}} \right)={{d}_{1}}+\lambda {{d}_{2}}\) represents a plane π₃ which is such that if any vector satisfies both the equation π₁ and π₂, it also satisfies the equation π₃. Cartesian Form: In cartesian system, let \({{\vec{n}}_{1}}={{A}_{1}}\hat{i}+{{B}_{1}}\hat{j}+{{C}_{1}}\hat{k}\), \({{\vec{n}}_{2}}={{A}_{2}}\hat{i}+{{B}_{2}}\hat{j}+{{C}_{2}}\hat{k}\) and \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\). Then (1) becomes \(x\left( {{A}{1}}+{{A}{2}}\lambda \right)+y\left( \(\left( {{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z-{{d}_{1}} \right)+\lambda \left( {{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z-{{d}_{2}} \right)=0\). Which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ. Example: Find the equation of a plane containing the line of intersection of the plane x + y+ z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through (1, 1, 1). Solution: The equation of plane passing through the line of intersection of the given planes is (x + y + z – 6) + λ (2x + 3y + 4z + 5) = 0 . . . (1) If it is passes through (1, 1, 1), (1 + 1 + 1 – 6) + λ (2.1 + 3.1 + 4.1 + 5) = 0 -3 + λ (14) = 0 λ = 3/14 λ value substitute in equation (1) (x + y + z – 6) + (3/14) (2x + 3y + 4z + 5) = 0 Required equation is 20x + 23y + 26z – 69 = 0.
Yes, you could convert equation (i) to parametric vector form, substitute it into (ii), make one parameter the subject, substitute it back into that parametric vector equation to finally obtain the intersection line's parametric vector equation. Notice that this doesn't actually involve finding $\vec a,$ which can be determined as the next step. A more direct method (also not requiring finding $\vec a)$ is illustrated by this example: \begin{gather}\vec{r} \cdot \begin{pmatrix} 7\cr2 \cr -3\end{pmatrix}=4 \tag{$\pi_1$}\\\vec{r} \cdot \begin{pmatrix} 2\cr1 \cr 0\end{pmatrix}=5\tag{$\pi_2$} \end{gather} $\pi_1:\;7x+2y-3z=4\\\pi_2:\;2x+y=5$ Expressing $x$ and $y$ in terms of $z$ $(\pi_1{-}2\pi_2,\,\ldots)$ gives the intersection line of $\pi_1$ and $\pi_2: \\\vec r=\begin{pmatrix} x\cr y \cr z\end{pmatrix}\\=\begin{pmatrix} z-2\cr 9-2z \cr z\end{pmatrix}\\=\begin{pmatrix} -2\cr 9 \cr 0\end{pmatrix}+z\begin{pmatrix} 1 \cr -2 \cr 1\end{pmatrix}\quad\left(z\in\mathbb R\right).$ |
The vector equation of the plane passing through the line of intersection of two planes
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