Problem 1
Question
Find the intercepts and sketch the graph of the plane. $$ 4 x+2 y+6 z=12 $$
Step-by-Step Solution
Verified Answer
The intercepts of the plane are (3, 0, 0), (0, 6, 0) and (0, 0, 2). The graph is a plane that intersects these points.
1Step 1: Find the x-intercept
Set y and z to zero, then solve the equation for x. This results in: \(4x = 12\), so \(x = 3\). The x-intercept is therefore the point (3, 0, 0).
2Step 2: Find the y-intercept
Set x and z to zero, then solve the equation for y. This leads to: \(2y = 12\), so \(y = 6\). The y-intercept is therefore the point (0, 6, 0).
3Step 3: Find the z-intercept
Set x and y to zero, then solve the equation for z. This gives: \(6z = 12\), so \(z = 2\). The z-intercept is therefore the point (0, 0, 2).
4Step 4: Sketch the graph
On a set of three-dimensional axes, plot the intercepts found in steps 1-3. Connect these points to form a triangle, then extend the edges to indicate the plane. The plane is the entire flat surface that these points define.
Other exercises in this chapter
Problem 1
Find any critical points and relative extrema of the function. $$ f(x, y)=x^{2}-y^{2}+4 x-8 y-11 $$
View solution Problem 1
Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ z=3 x+5 y-1 $$
View solution Problem 1
Plot the points on the same threedimensional coordinate system. (a) \((2,1,3)\) (b) \((-1,2,1)\)
View solution Problem 2
Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{3} \int_{0}^{1}(2 x+6 y) d y d x $$
View solution