What is the equation of a plane passing through the point (1, 2, 3) with normal

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Question: What is the equation of a plane passing through the point (1, 2, 3) with normal vector (2, -1, 3)? (2021)

Options:

Correct Answer: 2x - y + 3z = 0

Exam Year: 2021

Solution:

Equation of the plane: 2(x-1) - 1(y-2) + 3(z-3) = 0 simplifies to 2x - y + 3z = 12.

What is the equation of a plane passing through the point (1, 2, 3) with normal vector (2, -1, 3)? (2021)

What is the equation of a plane passing through the point (1, 2, 3) with normal vector (2, -1, 3)? (2021)

Step 1: Identify the point through which the plane passes. The point is (1, 2, 3).
Step 2: Identify the normal vector of the plane. The normal vector is (2, -1, 3).
Step 3: Use the point-normal form of the equation of a plane, which is given by: a(x - x0) + b(y - y0) + c(z - z0) = 0, where (x0, y0, z0) is the point and (a, b, c) is the normal vector.
Step 4: Substitute the values into the equation. Here, a = 2, b = -1, c = 3, and (x0, y0, z0) = (1, 2, 3).
Step 5: The equation becomes: 2(x - 1) - 1(y - 2) + 3(z - 3) = 0.
Step 6: Expand the equation: 2x - 2 - y + 2 + 3z - 9 = 0.
Step 7: Combine like terms: 2x - y + 3z - 9 - 2 + 2 = 0, which simplifies to 2x - y + 3z - 9 = 0.
Step 8: Rearrange the equation to the standard form: 2x - y + 3z = 9.

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