Find the equation of the plane passing through the points $(2, 3, - 4)$ and $(1, - 1, 3)$ and parallel to the $x - a x i s$ .

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Solution

A vector in the plane is
$\to u = ⎛ ⎜ ⎝ \begin{matrix} 1 - 1 3 \end{matrix} ⎞ ⎟ ⎠ - ⎛ ⎜ ⎝ \begin{matrix} 23 - 4 \end{matrix} ⎞ ⎟ ⎠ = ⎛ ⎜ ⎝ \begin{matrix} - 1 - 4 7 \end{matrix} ⎞ ⎟ ⎠$
A vector direction for the X-axis is
$\to v = ⎛ ⎜ ⎝ \begin{matrix} 100 \end{matrix} ⎞ ⎟ ⎠$
So,
A normal vector to the plane is
$∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 1 & - 4 & 7 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣^i (0 - 0) -^j (0 - 7) +^k (0 + 4) ⟨ 0, 7, 4 ⟩$
The equation of the plane is
$\begin{matrix} 0 (x - 2) + 7 (y - 3) + 4 (z + 4) = 0 7 y - 21 + 4 z + 16 = 0 7 y + 4 z - 5 = 0 \end{matrix}$

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