Question

Give the equation of the described plane in standard and general forms. Contains the point (2,-6,1) and the line $$ \ell(t)=\left\\{\begin{array}{l} x=2+5 t \\ y=2+2 t \\ z=-1+2 t \end{array}\right. $$

Short answer

The plane equation is \(2x - y - 4z = -17\).

Step-by-step solution

Understand the Problem

We are tasked with finding the equation of a plane which contains a given point (2, -6, 1) and a line defined by parametric equations in terms of a parameter \( t \). This line is given as: \( x = 2 + 5t \), \( y = 2 + 2t \), \( z = -1 + 2t \).

Determine a Point on the Line

The line itself provides infinitely many points, any can be used to verify the plane. Let's use \( t = 0 \) to find a specific point: when \( t = 0 \), \( x = 2 \), \( y = 2 \), and \( z = -1 \). So a point on the line is (2, 2, -1).

Find a Direction Vector of the Line

A direction vector for the line can be extracted from its parametric form, which is derived directly from the coefficients of \( t \). Here, the direction vector is \( \langle 5, 2, 2 \rangle \).

Establish Two Vectors in the Plane

We need two non-parallel vectors in the plane. We have a point (2, -6, 1) that is given. Use this and another point from the line, such as (2, 2, -1). Thus, vector from (2, -6, 1) to (2, 2, -1) is \( \langle 0, 8, -2 \rangle \). The second vector is the direction vector from the line, \( \langle 5, 2, 2 \rangle \).

Calculate the Cross Product

The normal vector to the plane can be found by computing the cross product of \( \langle 0, 8, -2 \rangle \) and \( \langle 5, 2, 2 \rangle \).\[ \vec{n} = \langle 0, 8, -2 \rangle \times \langle 5, 2, 2 \rangle = \langle (8 \cdot 2 - (-2) \cdot 2), ((-2) \cdot 5 - 0 \cdot 2), (0 \cdot 2 - 8 \cdot 5) \rangle = \langle 20, -10, -40 \rangle \].

Simplify Normal Vector

The normal vector \( \langle 20, -10, -40 \rangle \) can be simplified by dividing by 10: \( \langle 2, -1, -4 \rangle \).

Write Plane Equation in Standard Form

A plane equation from its normal vector \( \langle a, b, c \rangle \) and a point \((x_0, y_0, z_0)\) is \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \).Using point (2, -6, 1), the equation becomes: \[ 2(x-2) - 1(y+6) - 4(z-1) = 0 \].

Simplify to General Form

Expand and simplify the equation:\[ 2x - 1y - 4z = -17 \].This is the general form of the plane equation.

Key concepts

Cross Product

The concept of the cross product is crucial when working with 3D vectors, especially in relation to planes. The cross product of two vectors results in a third vector that is orthogonal, or perpendicular, to the plane containing the original vectors.

For vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \vec{a} \times \vec{b} \) results in a new vector \( \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \).

This operation is useful for finding a vector normal to a plane because, if you have two vectors lying on the plane, their cross product will point directly out of it.

• Increases understanding of orthogonal vectors.
• Essential for determining plane orientations.

Direction Vector

In the context of lines, the direction vector plays an important role in defining how the line extends through space. The direction vector is derived from the coefficients of the parameter in the line's parametric equations.

For a line given by \( x = a + pt \), \( y = b + qt \), \( z = c + rt \), the direction vector is \( \langle p, q, r \rangle \).

In our exercise, the line is \( x = 2 + 5t \), \( y = 2 + 2t \), \( z = -1 + 2t \), and the corresponding direction vector is \( \langle 5, 2, 2 \rangle \). This tells us how the line is oriented in space.

• Contains the line's slope information.
• Helps derive vectors within a plane.

Parametric Equations

Parametric equations allow us to describe the coordinates of a line in terms of a single variable, often denoting time or some other parameter.

These equations take the form of \( x = a + pt \), \( y = b + qt \), \( z = c + rt \), where \( a, b, c \) denote a point on the line, and \( \langle p, q, r \rangle \) is the direction vector.

By choosing different values for \( t \), we calculate different points on the line. This representation is particularly handy when working with points on lines as part of larger calculations, such as determining a plane.

They provide insight into the geometric progression of the line in 3D space, demonstrating how a line traverses through points.

• Useful for finding specific line points.
• Simplifies complex geometric calculations.

Normal Vector

The normal vector is critical for defining a plane uniquely in 3-dimensional space. It acts like an arrow that points directly out of the plane, perpendicular to it.

Derived by taking the cross product of two non-parallel vectors lying in the plane, the normal vector ensures the plane's orientation is clearly defined.

In plane equation form, if the normal vector is \( \langle a, b, c \rangle \), it reflects the coefficients used in the standard plane equation format: \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \).

For our problem, we found the normal vector \( \langle 2, -1, -4 \rangle \) by simplifying \( \langle 20, -10, -40 \rangle \), which was calculated using the cross product. This facilitates the translation of plane position and orientation into an equation format.

• Indicates plane orientation and steepness.
• Links geometric properties with algebraic expressions.