Can anyone please help me to solve this problem? Please I am stuck! **Title: Finding an Equation of a Plane Orthogonal to a Line****Introduction:**In this exercise, we will determine the equation of a plane that passes through a given point and is orthogonal to a specified line. This requires understanding the relationship between planes and lines in three-dimensional space.**Problem:**Find an equation of a plane through the point \( (-5, 4, -3) \) which is orthogonal to the line given by the parametric equations:- \( x = 5t - 1 \)- \( y = -8t + 5 \)- \( z = -1 + 3t \)**Solution Outline:**1. **Determine the Direction Vector:** The line's parametric equations provide the direction vector of the line. For the given line, the direction vector is \( \langle 5, -8, 3 \rangle \).2. **Use the Direction Vector as the Normal Vector:** Since the plane is orthogonal to the line, the direction vector of the line will be the normal vector of the plane.3. **Point-Normal Form of the Plane Equation:** The general equation of a plane in point-normal form is: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \] where \( \langle a, b, c \rangle \) is the normal vector and \( (x_0, y_0, z_0) \) is a point on the plane.4. **Substitute the Values:** - Normal Vector: \( \langle 5, -8, 3 \rangle \) - Point on the Plane: \( (-5, 4, -3) \) The plane equation becomes: \[ 5(x + 5) - 8(y - 4) + 3(z + 3) = 0 \]5. **Simplify the Equation:** \[ 5x + 25 - 8y + 32 + 3z + 9 = 0 \] \[ 5x - 8y + 3z + 66 = 0 \]**Conclusion:**The equation of the plane orthogonal to the given line and passing through **Problem:**Find an equation of a plane through the point \((-5, 4, -3)\) which is orthogonal to the line:\[ x = 5t - 1 \]\[ y = -6t + 4 \]\[ z = -1(t + 3t) \]**Solution Explanation:**To find the equation of a plane that passes through a given point and is orthogonal to a line, you need to determine the direction vector of the line and use it as the normal vector for the plane.1. **Find the direction vector of the line**: - From the parametric equations of the line, the coefficients of \(t\) in each equation represent the direction vector. So, the direction vector \(\mathbf{d}\) is \((5, -6, -1)\).2. **Use the direction vector as the normal vector of the plane**: - The general equation of a plane with a normal vector \((a, b, c)\) passing through a point \((x_0, y_0, z_0)\) is: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]3. **Substitute the given point and normal vector into the plane equation**: - For the point \((-5, 4, -3)\) and normal vector \((5, -6, -1)\): \[ 5(x + 5) - 6(y - 4) - 1(z + 3) = 0 \]4. **Simplify the equation to find the plane through the given point**: - Expand and simplify the equation: \[ 5x + 25 - 6y + 24 - z - 3 = 0 \] \[ 5x - 6y - z + 46 = 0 \]Thus, the equation of the plane is \(5x - 6y - z + 46 = 0\).

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Answered: Find an equation of a plane through the…

Find an equation of a plane through the point C-5, 4₁-32 Which is or thogonal to the line X= 5+-1₁ y=-30 +4+), Z==(1+3+)

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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