The equation of the plane passing through $(2, 3, 1)$ , $(5, - 2, 1)$ and parallel to $z - a x i s$ is

5 x + 3 y = 19

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5 x + 3 y + 19 = 0

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3 x + 5 y = 21

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3 x - 5 y + 9 = 0

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Solution

The correct option is A $5 x + 3 y = 19$
$A (2, 3, 1), B (5, - 2, 1)$
Plane is parallel to z-axis $=$ normal is $⊥$ to z-axis.
$- - \to A B = 3^i - 5^j + 0^k$
$\to n$ be normal vector.
$∴ \to n = - - \to A B \times (R) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 3 & - 5 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 5^i - 3^j$
$∴$ directions of normal are $(5, 3, 0)$
Equation, $5 x + 3 y = 19$

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Similar questions

3 x + 5 y = 19

(3, 2)

The equation of the line passing through $(1, 5)$ and perpendicular to the line $3 x - 5 y + 7 = 0$ is

5x-3y=1

2x+5y-19=0

3 x - 5 y - 19 = 0, - 7 x + 3 y + 1 = 0 .

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