Question

Find a point on the surface \(x^{2}+2 y^{2}+3 z^{2}=12\) where the tangent plane is perpendicular to the line with parametric equations: \(x=1+2 t, y=3+8 t, z=2-6 t\)

Short answer

The points are (1, 2, -1) and (-1, -2, 1).

Step-by-step solution

Identify the Gradient of the Surface

For the surface given by the equation \(x^2 + 2y^2 + 3z^2 = 12\), we first find the gradient, which is a vector of partial derivatives with respect to each variable: \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = \left( 2x, 4y, 6z \right)\).

Find the Direction Vector of the Line

The direction vector of the line described by the parametric equations \(x=1+2t, y=3+8t, z=2-6t\) is \(\mathbf{v} = (2, 8, -6)\).

Set Gradient Equal to Line Direction Vector

For the tangent plane to be perpendicular to the line, the gradient of the surface at a point must be parallel to the direction vector of the line. Thus, we set \(abla f = k\mathbf{v}\), where \(k\) is a scalar. This yields the system of equations \(2x = 2k\), \(4y = 8k\), \(6z = -6k\).

Solve for Scalar and Coordinates

Solving the equations \(2x = 2k\), \(4y = 8k\), \(6z = -6k\), we find \(x = k\), \(y = 2k\), \(z = -k\).

Substitute in Surface Equation

Substitute \(x = k\), \(y = 2k\), \(z = -k\) into the surface equation \(x^2 + 2y^2 + 3z^2 = 12\):\[k^2 + 2(2k)^2 + 3(-k)^2 = 12\]This simplifies to \[k^2 + 8k^2 + 3k^2 = 12\], which gives \(12k^2 = 12\).

Solve for k

Solving \(12k^2 = 12\), we get \(k^2 = 1\), hence \(k = 1\) or \(k = -1\).

Find the Points

Substituting the values of \(k\) back into the expressions for \(x, y, z\): For \(k = 1\): \(x = 1\), \(y = 2\), \(z = -1\). For \(k = -1\): \(x = -1\), \(y = -2\), \(z = 1\).

Key concepts

Parametric Equations

Understanding parametric equations is essential when dealing with lines in three-dimensional space. A parametric equation expresses each coordinate in terms of a single parameter, often denoted by \( t \). For a line, the parametric equations can be written as:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

where \((x_0, y_0, z_0)\) is a point on the line, and \( a, b, c \) are the components of the direction vector of the line.

In the given exercise, the line's parametric equations are \( x=1+2t \), \( y=3+8t \), and \( z=2-6t \). This means the line passes through the point \((1, 3, 2)\) and moves according to the direction vector \((2, 8, -6)\). Understanding these equations helps visualize the line in space and determine how it relates to other geometric entities like planes and surfaces.

Direction Vector

The direction vector is crucial for understanding the orientation of a line in space. It represents how the line extends from a given point and is derived from the parametric equations.

From the parametric equations \( x=1+2t \), \( y=3+8t \), \( z=2-6t \), we extract the direction vector \( \mathbf{v} = (2, 8, -6) \).

This vector indicates that for every unit increase in \( t \), the line moves 2 units in the x-direction, 8 units in the y-direction, and -6 units in the z-direction. It plays a significant role in parallelism and perpendicularity discussions. In this exercise, to find when a tangent plane is perpendicular to the line, the gradient of the surface must align with this direction vector.

Surface Equation

The surface equation details the mathematical representation of a three-dimensional surface. For example, the equation \( x^2 + 2y^2 + 3z^2 = 12 \) describes an ellipsoidal surface in three-dimensional space.

Every point on this surface satisfies the equation, meaning that if you plug in the coordinates of any point into \( x, y, z \), the equation holds true. The surface's shape depends on the coefficients of each term, which in this case create an elongated shape around the axes due to the differing coefficients (1, 2, and 3).

Recognizing this equation helps determine points, like those from which the tangent plane is perpendicular to a given line, by using gradients and other methods in more complex calculations.

Partial Derivatives

Partial derivatives help in understanding how a function changes as one of the variables changes, with others held constant. For a surface given by \( f(x, y, z) = 12 \), the gradient vector is composed of partial derivatives:

• \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \)

For the equation \( x^2 + 2y^2 + 3z^2 = 12 \), the partial derivatives are \( 2x, 4y, \) and \( 6z \), giving us the gradient vector \( (2x, 4y, 6z) \).

The gradient represents the direction of steepest ascent on the surface at a given point. In this particular problem, the gradient's role is crucial in finding a point on the surface where the tangent plane is perpendicular to a line by ensuring the gradient is parallel to the line's direction vector.