Question

Find the equation of the plane through the given points. $$ (a, 0,0),(0, b, 0) \text { , and }(0,0, c), \text { (None of } a, b \text { , and } c \text { is zero.) } $$

Short answer

The plane equation is \(bcx + acy + abz = abc\).

Step-by-step solution

Identify the Normal Vector

Find a vector that is normal (perpendicular) to the plane. The points given are \((a, 0, 0), (0, b, 0), (0, 0, c)\). We need to find two vectors in the plane. These are \(\vec{AB} = (0 - a, b - 0, 0 - 0) = (-a, b, 0)\) and \(\vec{AC} = (0 - a, 0 - 0, c - 0) = (-a, 0, c)\). The normal vector \(\vec{n}\) can be found by taking the cross product of these vectors.

Calculate the Cross Product

Calculate the cross product \(\vec{AB} \times \vec{AC}\) to find the normal vector. We use the determinant method:\[\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -a & b & 0 \ -a & 0 & c \end{vmatrix} = \left(bc\right)\hat{i} + \left(ac\right)\hat{j} + \left(ab\right)\hat{k}\]Thus, the normal vector is \(\vec{n} = (bc, ac, ab)\).

Write the Plane Equation

Use the normal vector and one of the given points (for example, \((a, 0, 0)\)) to write the equation of the plane. The general equation of a plane is given by \(Ax + By + Cz = D\) where \((A, B, C)\) is the normal vector. So, we have:\[bc(x - a) + ac(y - 0) + ab(z - 0) = 0\]Simplifying, the equation of the plane can be written as:\[bcx + acy + abz = abc\]

Simplify the Plane Equation

Ensure the equation is simplified by arranging the terms. The equation \(bcx + acy + abz = abc\) is already simplified. This is a standard form for the equation of the plane.

Key concepts

Normal vector

A normal vector is crucial in determining the unique orientation of a plane. In three-dimensional space, a normal vector is perpendicular to the plane and helps in formulating its equation. When dealing with planes, identifying a normal vector gives us a shortcut to knowing how the plane is aligned or tilted.
To find a normal vector to a plane through specific points, first identify vectors that lie within the plane. For the plane through points a. ext(0,b,0) a. (c), and we have vectors a x b a * (0,0 a, b,0) a). Their cross product reveals a vector that stands at right angles to the plane's surface.

• Normal vectors make understanding spatial relationships easier.
• Essential in geometry for aligning and tilting planes.

Cross product

The cross product is a mathematical operation applied to two vectors in space to find a third vector that is perpendicular to both. This new vector is vital in finding the normal vector of a given plane.
For vectors a, / b (=a, a) and to .apply a. aB C, we apply the cross product formula. The result is another vector which is normal to the two original ones, a the orientation bc, .ac ), distinct ac the given. This helps in determining the plane's equation, ensuring it's arithmetic and spatially correct.

• Provides a normal vector from in-plane vectors.
• Useful in physics, engineering, and computer graphics to define rotations or orientations.

Determinant method

The determinant method is a structured way to compute the cross product using a matrix set-up. This technique is efficient for finding the normal vector when given two vectors within a plane.
To use the determinant method, arrange the unit vectors (units in three-dimensional vector space:i, j, and k a.a.in theplane, abvectora. and ab , etc.). As shown in our example, it's organized as:\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -a & b & 0 \ -a & 0 & c \end{vmatrix} \)

Simplifies complex vector calculations.

Standard in math and science for ensuring accurate cross product computation.

Simplified plane equation

Crafting a plane equation involves using the components of the normal vector and a point on the plane. A simplified equation presents a clear and easy-to-handle form for computations and interpretations.
For our exercise, after calculating the normal vector (abcomponents,oangle:, acontributecalculate ac), (aorbc), and determine ).where (in a general form: \[ Ax + By + Cz = D \]. We see this represents the relationship with any plane point. It's crucial to further expand and simplify the components until you reach an explicit equation.
This allows everyone, from students to professionals, to directly see the plane's orientation in space.

• Makes equations more readable and interpretable.
• Useful across many fields including mathematics, physics, and engineering.
• Leads to better understanding of geometric and algebraic relationships.