The Planes — Flashcards

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The general equation of a plane passing through a point $(x_1, y_1, z_1)$ is: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ Here, $x_1 = 2$, $y_1 = -1$, $z_1 = 3$ So, the equation is: $a(x - 2) + b(y…

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Let the points be: A(1, 2, 3), B(2, 3, 4), C(3, 4, 5) Using the determinant formula: $$ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_…

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Let: $P_1 = (1,1,1)$, $P_2 = (2,3,4)$, $P_3 = (3,2,1)$ Use determinant: $$ \begin{vmatrix} x - 1 & y - 1 & z - 1 \\ 2 - 1 & 3 - 1 & 4 - 1 \\ 3 - 1 & 2 - 1 & 1 - 1 \end{vmatrix} = 0 \Rightarrow \begi…

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Given: $3x + 4y - 5z = 30$ Divide both sides by 30: $$ \frac{3x}{30} + \frac{4y}{30} - \frac{5z}{30} = 1 \Rightarrow \frac{x}{10} + \frac{y}{7.5} + \frac{z}{-6} = 1 $$ So, intercepts are: - x-inter…

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To find intercepts, set two variables to zero: x-intercept: Set $y = 0$, $z = 0$ $2x = 12 \Rightarrow x = 6$ y-intercept: Set $x = 0$, $z = 0$ $3y = 12 \Rightarrow y = 4$ z-intercept: Set $x = 0$, …

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Given: $2x - 3y + 6z = 6$ Magnitude of normal vector: $\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ Divide both sides by 7: $$ \frac{2}{7}x - \frac{3}{7}y + \frac{6}{7}z = \frac{6…

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The foot of the perpendicular is (2, -1, 2), so the normal vector to the plane is $\vec{n} = (2, -1, 2)$ The plane passes through (2, -1, 2) So, equation is: $2(x - 2) - 1(y + 1) + 2(z - 2) = 0$ $…

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For two planes: $\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$ Here: Plane 1: $a_1 = 2$, $b_1 = 1$, $c_1 = -2$ Plane 2: $a_2 = 3$, $…