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Find the equation of a plane through the points \( (2,1,2),(6,-6,-2) ;(4,6,3) \). Question Help: \( \square \) Video

Ask by King Bowers. in the United States
Jan 13,2025

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The equation of the plane is \(13x - 12y + 34z = 82\).

Solution

To find the equation of the plane passing through the points \( P_1 = (2, 1, 2) \), \( P_2 = (6, -6, -2) \), and \( P_3 = (4, 6, 3) \), follow these steps: ### 1. Find Two Vectors in the Plane First, determine two vectors that lie on the plane by subtracting the coordinates of the points: \[ \vec{u} = P_2 - P_1 = (6-2, -6-1, -2-2) = (4, -7, -4) \] \[ \vec{v} = P_3 - P_1 = (4-2, 6-1, 3-2) = (2, 5, 1) \] ### 2. Compute the Normal Vector The normal vector \(\vec{n}\) to the plane can be found by taking the cross product of \(\vec{u}\) and \(\vec{v}\): \[ \vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -7 & -4 \\ 2 & 5 & 1 \\ \end{vmatrix} \] Calculating the determinant: \[ \vec{n} = \left( (-7)(1) - (-4)(5) \right) \mathbf{i} - \left( 4(1) - (-4)(2) \right) \mathbf{j} + \left( 4(5) - (-7)(2) \right) \mathbf{k} \] \[ \vec{n} = ( -7 + 20 )\mathbf{i} - (4 + 8)\mathbf{j} + (20 + 14)\mathbf{k} \] \[ \vec{n} = 13\mathbf{i} - 12\mathbf{j} + 34\mathbf{k} \] So, the normal vector is \( \vec{n} = (13, -12, 34) \). ### 3. Write the Equation of the Plane Using the point-normal form of the plane equation: \[ 13(x - 2) - 12(y - 1) + 34(z - 2) = 0 \] Expanding and simplifying: \[ 13x - 26 - 12y + 12 + 34z - 68 = 0 \] \[ 13x - 12y + 34z - 82 = 0 \] ### Final Equation \[ 13x - 12y + 34z = 82 \] **Answer:** After simplifying, an equation for the plane is 13 x – 12 y + 34 z = 82.

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To find the equation of the plane that passes through the points \( A(2, 1, 2) \), \( B(6, -6, -2) \), and \( C(4, 6, 3) \), we first need to determine two vectors that lie on the plane. We can use vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \): 1. Calculate \( \overrightarrow{AB} = B - A = (6 - 2, -6 - 1, -2 - 2) = (4, -7, -4) \) 2. Calculate \( \overrightarrow{AC} = C - A = (4 - 2, 6 - 1, 3 - 2) = (2, 5, 1) \) Next, we find the normal vector \( \mathbf{N} \) by taking the cross product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \): \[ \mathbf{N} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -7 & -4 \\ 2 & 5 & 1 \end{vmatrix} \] Calculating the determinant: \[ \mathbf{N} = \mathbf{i}((-7)(1) - (-4)(5)) - \mathbf{j}((4)(1) - (-4)(2)) + \mathbf{k}((4)(5) - (-7)(2)) \] \[ \mathbf{N} = \mathbf{i}(-7 + 20) - \mathbf{j}(4 + 8) + \mathbf{k}(20 + 14) \] \[ \mathbf{N} = \mathbf{i}(13) - \mathbf{j}(12) + \mathbf{k}(34) = (13, -12, 34) \] Now, we can use the normal vector and a point to form the plane's equation. Using point \( A(2, 1, 2) \): The equation of the plane is given by \( N_x(x - x_0) + N_y(y - y_0) + N_z(z - z_0) = 0 \): \[ 13(x - 2) - 12(y - 1) + 34(z - 2) = 0 \] Expanding this gives: \[ 13x - 26 - 12y + 12 + 34z - 68 = 0 \] Simplifying further: \[ 13x - 12y + 34z - 82 = 0 \] So the equation of the plane is: \[ 13x - 12y + 34z = 82 \] Now, let’s sprinkle some extra knowledge! Did you know that planes in three-dimensional space are defined using a point and a normal vector? The normal vector is crucial because it tells you the orientation of the plane in space. If you visualize a plane, it’s like a sheet of paper – the normal vector would be a stick that pokes straight out from the paper! When solving for the plane equation, a common mistake is calculating the cross product incorrectly. Always remember to set up your determinant correctly! A little tip? Double-check your signs when expanding the determinant; it can save you some headaches later on!

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