Problem 23 - Precalculus Mathematics for Calculus

Question

Find an equation of the plane that passes through the points \(P, Q,\) and \(R\) $$P\left(3, \frac{1}{3},-5\right), \quad Q\left(4, \frac{2}{3},-3\right), \quad R(2,0,1)$$

Step-by-Step Solution

Answer

The equation of the plane is \(3x - 12y = 5\).

Step 1: Find Vectors from Points

To find the equation of a plane through three points, we first need vectors that lie on the plane. We can form two vectors by subtracting the coordinates of the given points:\( \vec{PQ} = \langle 4-3, \frac{2}{3}-\frac{1}{3}, -3-(-5) \rangle = \langle 1, \frac{1}{3}, 2 \rangle \)\( \vec{PR} = \langle 2-3, 0-\frac{1}{3}, 1-(-5) \rangle = \langle -1, -\frac{1}{3}, 6 \rangle \)

Step 2: Find the Normal Vector using Cross Product

The cross product of \(\vec{PQ}\) and \(\vec{PR}\) gives a vector normal to the plane. Compute the cross product:\[ \vec{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & \frac{1}{3} & 2 \ -1 & -\frac{1}{3} & 6 \end{vmatrix} \]Compute the determinant:\[ \vec{n} = \left(\left(\frac{1}{3} \cdot 6\right) - \left(-\frac{1}{3} \cdot 2\right)\right)\mathbf{i} - \left(\left(1 \cdot 6\right) - \left(-1 \cdot 2\right)\right)\mathbf{j} + \left(1 \cdot (-\frac{1}{3}) - \left(\frac{1}{3} \cdot -1\right)\right)\mathbf{k} \]Simplify to find:\[ \vec{n} = \langle 2, -8, 0 \rangle \]

Step 3: Formulate the Plane Equation

With the normal vector \(\vec{n} = \langle 2, -8, 0 \rangle \), use the point \(P(3, \frac{1}{3}, -5)\) to write the plane equation in the form \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \):\[ 2(x-3) - 8\left(y-\frac{1}{3}\right) + 0(z+5) = 0 \]Simplify to:\[ 2x - 8y + \frac{8}{3} = 6 \]Simplifying further gives:\[ 6x - 24y + 8 = 18 \]Therefore, the equation becomes:\[ 6x - 24y = 10 \]

Step 4: Simplify the Equation

Further simplify the equation by factoring out common terms:\[ 6x - 24y = 10 \]Divide through by 2:\[ 3x - 12y = 5 \]

Key Concepts

Cross Product

To understand the cross product, imagine you need to find a vector that is perpendicular to a plane defined by two vectors. The cross product allows you to compute this perpendicular (or normal) vector. If you have two vectors \(\vec{a}\) and \(\vec{b}\), the cross product is \(\vec{a} \times \vec{b}\) and results in a third vector \(\vec{n}\), which is orthogonal to both \(\vec{a}\) and \(\vec{b}\).

This concept can be visualized with your right hand: if you point your index finger along \(\vec{a}\) and your middle finger along \(\vec{b}\), your thumb points in the direction of \(\vec{a} \times \vec{b}\). The formula for computing the cross product in three dimensions uses a determinant:

• \(\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\)

This can be further expanded using minors to find each component of the resulting vector \(\vec{n}\).

It's important because finding a perpendicular vector forms the basis for defining planes, making the cross product powerful in geometry and physics.

Normal Vector

Once you've determined the cross product of two vectors that lie on a plane, you've found what's called the normal vector to that plane. The normal vector is crucial because it's perpendicular to every vector that exists on the plane. Essentially, it gives you a way to talk about the 'direction' of the plane even though a plane itself doesn't move in one direction.

For the plane through points \(P\), \(Q\), and \(R\), you find two vectors like \(\vec{PQ}\) and \(\vec{PR}\). The cross product of \(\vec{PQ}\) and \(\vec{PR}\) gives you the normal vector. In this exercise:

• The cross product results in a normal vector \(\vec{n} = \langle 2, -8, 0 \rangle\).

This normal vector has components \(a\), \(b\), and \(c\), which are critical for forming the equation of the plane. It essentially helps define the orientation of the plane in space.

Plane Through Points

To find a plane through specific points, understanding the spatial relationship between those points is key. First, calculate two vectors that lie on the plane by subtracting the coordinates of the points in pairs. This results in vectors \(\vec{PQ}\) and \(\vec{PR}\), which physically extend from one point to another while laying on the plane surface.

The equation of a plane in space is generally given by \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the components of the normal vector. In our exercise, having the normal vector \(\vec{n} = \langle 2, -8, 0 \rangle\), and knowing it passes through point \(P(3, \frac{1}{3}, -5)\), allows us to write the plane equation, noting:

• You substitute \(x_0, y_0, z_0\) from point \(P\) and simplify the expression.

The simplified form becomes: \(3x - 12y = 5\). This equation reflects all solutions \((x, y, z)\) that fit our plane, including the three initial points.