Find the vector equation of the plane passing through the intersection of the planes $\to r \cdot (^i +^j +^k) = 6$ and $\to r \cdot (2^i + 3^j + 4^k) = - 5$ and the point $(1, 1, 1)$ .

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Solution

Here, ${\to n}_{1} =^i +^j +^k$ and ${\to n}_{2} =^2 i + 3^j + 4^k$
$d_{1} = 6$ and $d_{2} = - 5$
Hence, using the relation $\to r . ({\to n}_{1} + λ {\to n}_{2}) = d_{1} + λ d_{2}$ , we get

$\to r \cdot [(^i +^j +^k) + λ (^2 i + 3^j + 4^k)] = 6 - 5 λ$
$\Rightarrow \to r \cdot [(1 + 2 λ)^i + (1 + 3 λ)^j + (1 + 4 λ)^k] = 6 - 5 λ \dots (1)$
where, $λ$ is some real number.

Taking $\to r = x^i + y^j + z^k$ , we get
$(x^i + y^j + z^k) . [(1 + 2 λ)^i + (1 + 3 λ)^j + (1 + 4 λ)^k] = 6 - 5 λ$
$\Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z = 6 - 5 λ$
$\Rightarrow (x + y + z - 6) + λ (2 x + 3 y + 4 z + 5) = 0 \dots (2)$

Given that the plane passes through the point $(1, 1, 1)$ , it must satisfy equation(2), i.e.,
$(1 + 1 + 1 - 6) + λ (2 + 3 + 4 + 5) = 0$
$\Rightarrow - 3 + λ (14) = 0 \Rightarrow λ = \frac{3}{14}$

Putting the value of $λ$ in $(1)$ , we get

$\Rightarrow \to r \cdot [(1 + \frac{3}{7})^i + (1 + \frac{9}{14})^j + (1 + \frac{6}{7})^k] = 6 - \frac{15}{14}$
$\Rightarrow \to r \cdot (\frac{10}{7}^i + \frac{23}{14}^j + \frac{13}{7}^k) = \frac{69}{14}$
$\Rightarrow \to r \cdot (20^i + 23^j + 26^k) = 69$
Which is the required vector equation of the plane.

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Find the vector equation of the plane passing through the intersection of the planes →r⋅(^i+^j+^k)=6 and →r⋅(2^i+3^j+4^k)=−5 and the point (1,1,1).

Find the vector equation of the plane passing through the intersection of the planes $\to r \cdot (^i +^j +^k) = 6$ and $\to r \cdot (2^i + 3^j + 4^k) = - 5$ and the point $(1, 1, 1)$ .