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 (1, 1, 1)? (2022)

Options:

Correct Answer: x + y + z = 6

Exam Year: 2022

Solution:

Equation of the plane: 1(x-1) + 1(y-2) + 1(z-3) = 0 => x + y + z = 6.

What is the equation of a plane passing through the point (1, 2, 3) with normal vector (1, 1, 1)? (2022)

What is the equation of a plane passing through the point (1, 2, 3) with normal vector (1, 1, 1)? (2022)

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 (1, 1, 1).
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) are the components of the normal vector.
Step 4: Substitute the values into the equation. Here, a = 1, b = 1, c = 1, x0 = 1, y0 = 2, z0 = 3.
Step 5: The equation becomes: 1(x - 1) + 1(y - 2) + 1(z - 3) = 0.
Step 6: Simplify the equation: (x - 1) + (y - 2) + (z - 3) = 0.
Step 7: Combine like terms: x + y + z - 6 = 0.
Step 8: Rearrange the equation to standard form: x + y + z = 6.

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