Question

The equation of plane passing through point \((1,-1,2)\) and \(\underline{r}=(1,1,1)+\mathrm{k}(2,1,2), \mathrm{k} \in \mathrm{R}\) is (A) \(5 \mathrm{x}-2 \mathrm{y}-4 \mathrm{z}+1=0\) (B) \(5 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}+1=0\) (C) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+1=0\) (D) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}=1\)

Short answer

The short answer is: \(-2y + 5x = 1\), so the correct option is (D) \(5x - 2y + 4z = 1\).

Step-by-step solution

Vector AB (from A to B)

To find the vector AB, we need to subtract the coordinates of A from the coordinates of B, which are given by: \(x = 1+k(2), y = 1+k(1), z = 1+k(2)\). So, AB = (x-1, y+1, z-2) = (2k, k, 2k). -Step 2 - Finding a normal vector to the plane The normal vector to the plane can be found by taking the cross product of two vectors in the plane.

Cross product of vector AB and direction vector of the line

We have the direction vector of the line as D = (2, 1, 2) and the vector AB = (2k, k, 2k). The cross product of these two vectors is given by: N = AB x D Using the determinant formula, we have: N = \(| \begin{matrix}\ \textbf{i}&\textbf{j}&\textbf{k} \\\ 2k&k&2k \\\ 2&1&2 \end{matrix} | = k \textbf{i} - 2k \textbf{j} + 0 \textbf{k} \) As k is any real number, we can consider k = 1 for simplicity, so N = (1, -2, 0) -Step 3 - Finding the equation of the plane Now, we have a point A(1, -1, 2) and a normal vector N = (1, -2, 0) for the plane.

Point-normal form of the plane

Using the point-normal form for the equation of the plane, we have: \(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\), where (x1, y1, z1) is a point on the plane and (a, b, c) is a normal vector to the plane. Plugging in the values, we get: \(1(x - 1) - 2(y + 1) + 0(z - 2) = 0\) Simplifying the equation, we have: \(-2y + 5x = 1\) Comparing this equation with the provided options, we can conclude that the correct answer is (D) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}=1\)

Key concepts

Vector Algebra

Vector algebra is a crucial mathematical tool used in the study of geometry and physics. It involves various operations between vectors, which are quantities defined by both a magnitude and a direction.

Vectors can be added together or multiplied by scalars to produce new vectors. In the context of our exercise, vector subtraction is used to find the vector from point A to point B, noted as vector AB. This operation is performed by subtracting the coordinates of the starting point A from the coordinates of the endpoint B.

The beauty of vector algebra lies in its ability to simplify complex geometrical problems into algebraic ones. By representing points, lines, and planes through vectors, we can utilize algebraic methods to solve problems that would otherwise require intricate geometrical constructions.

Cross Product

The cross product, also known as the vector product, is an operation exclusive to three-dimensional space that takes two vectors and returns a third vector that is perpendicular to both.

The resulting vector's magnitude corresponds to the area of the parallelogram formed by the two input vectors as its sides. This perpendicular vector can be calculated using the determinant of a matrix composed of the unit vectors along the x, y, and z axes and the components of the two input vectors.

For instance, in our problem, the cross product is used to find a normal vector to the plane by taking the vector formed by a point on the line and the line's direction vector. The concept of the cross product is essential in defining the orientation and properties of planes in three-dimensional space.

Point-Normal Form

The point-normal form is one of the methods to define the equation of a plane in three dimensions. It utilizes a known point on the plane and a normal vector orthogonal to the plane. The general equation is given by:
\[a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\],
where \((x_1, y_1, z_1)\) are the coordinates of the known point, and \(a\), \(b\), and \(c\) are the components of the normal vector.

In the exercise, after finding the normal vector using the cross product, point-normal form is applied to establish the equation that represents the plane. The simplicity of this form allows for straightforward calculation and understanding of a plane's orientation in space.

Plane Geometry in 3D

Plane geometry in 3D, or three-dimensional geometry, involves the study of shapes like planes, lines, and points existing in three-dimensional space. A plane can be considered as a flat surface extending infinitely in two dimensions, characterized by any three points that are not in a straight line, or a normal vector and a point on the plane.

Our problem illustrates the concept of a plane defined by a line (through a given point and a direction vector) and another distinct point not on that line. Such planes can be expressed algebraically, and mastering the representation and interpretation of these planes is critical for solving higher-level geometry problems, computer graphics, architectural design, and other technical fields that involve spatial reasoning.