Math / Geometry

QuestionFind an equation of the plane with $x$ -intercept $a, y$ -intercept $b$ , and $z$ -intercept $c$ .

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a plane that crosses the $x$ , $y$ , and $z$ axes at the points $a$ , $b$ , and $c$ respectively. Watch out! Remember that an intercept is a point where the plane crosses an axis, not just a number!
So the $x$ -intercept is at the point $(a, 0, 0)$ , the $y$ -intercept is at $(0, b, 0)$ , and the $z$ -intercept is at $(0, 0, c)$ !

STEP 2

1. Find a normal vector
2. Build the equation

STEP 3

We have three points in the plane: $(a, 0, 0)$ , $(0, b, 0)$ , and $(0, 0, c)$ .
Let's use these to create two vectors that lie within the plane.
We can do this by finding the difference between pairs of points.
Vector 1: $(0, b, 0) - (a, 0, 0) = (-a, b, 0)$
Vector 2: $(0, 0, c) - (a, 0, 0) = (-a, 0, c)$

STEP 4

A vector that's perpendicular to both of these vectors will be normal to the plane!
We can find this normal vector by taking the cross product of the two vectors we just found.
$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix} = (b c)\mathbf{i} - (-a c)\mathbf{j} + (b a)\mathbf{k} = bc\mathbf{i} + ac\mathbf{j} + ab\mathbf{k}$ So, our normal vector is $(bc, ac, ab)$ !

STEP 5

The general equation of a plane is given by $Ax + By + Cz = D$ , where $(A, B, C)$ is a normal vector to the plane.
We just found a normal vector: $(bc, ac, ab)$ .
So we can plug those values in for $A$ , $B$ , and $C$ :
$bcx + acy + abz = D$

STEP 6

To find $D$ , we can use any point on the plane.
Let's use the x-intercept, $(a, 0, 0)$ :
$bc(a) + ac(0) + ab(0) = D$ $abc = D$

STEP 7

Now, substitute the value of $D$ back into the equation:
$bcx + acy + abz = abc$ Let's divide both sides of the equation by $abc$ (assuming $a$ , $b$ , and $c$ are non-zero):
$\frac{bcx}{abc} + \frac{acy}{abc} + \frac{abz}{abc} = \frac{abc}{abc}$ $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

STEP 8

The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ .

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