Find the vector equation of the plane through the points $(2, 1, - 1)$ and $(- 1, 3, 4)$ and perpendicular to the plane $x - 2 y + 4 z = 10$ .

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Solution

The required plane passes through points $P (2, 1, - 1)$ and $Q (- 1, 3, 4) .$
We can write the position vectors as follows :
$a_{1} = 2 \to i + \to j - \to k$ and $a_{2} = - \to i + 3 \to j + 4 \to k$
$\Rightarrow$ $P Q = (a_{2} - a_{1}) = - 3 \to i + 2 \to j + 5 \to k$

$_{i}$ is the normal vector of plane $x - 2 y + 4 z = 10$
Let $\to i$ be the normal vector to the desired plane $\to n =_{i} \times - - \to P Q = ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to i & \to j & \to k 1 & - 2 & 4 - 3 & 2 & 5 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow$ $\to n = \to i (- 10 - 8) - \to j (5 + 12) + \to k (2 - 6)$
$\Rightarrow$ $\to n = - 18 \to i - 17 \to j - 4 \to k$

$\Rightarrow$ $\to r . \to n = \to a . \to n$
$\Rightarrow$ $\to r . (- 18 \to i - 17 \to j - 4 \to k) = (2 \to i + \to j - \to k) (18 \to i - 17 \to j - 4 \to k)$
$\Rightarrow$ $\to r . (- 18 \to i - 17 \to j - 4 \to k) = - 36 - 17 + 4$
$\Rightarrow$ $\to r . (18 \to i - 17 \to j + 4 \to k) = 49$

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