Mathematics, 01.08.2019 02:30 bryn2433

Find the equation of the plane which has intercepts of (4,0,0) and (0,−2,0) and no z-intercept.

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Answer from: XVARYX

The equation of the plane which has intercepts of (4, 0, 0) and (0, -2, 0) and no z-intercept is $\frac{1}{4}\cdot x -\frac{1}{2}\cdot y = 1$ .

A plane is represented by the following expression:

$A\cdot x + B\cdot y + C\cdot z = 1$ (1)

Where A,B, C are real coefficients.

We need three distinct points to determine all coefficients of the plane. If we know that $(x_{1}, y_{1}, z_{1}) = (4,0,0)$ , $(x_{2}, y_{2}, z_{2}) = (0,-2,0)$ and $(x_{3}, y_{3}, z_{3}) = (0, 0, 0)$ , then we have the following system of linear equations:

$4\cdot A = 1$ (2)

$-2\cdot B = 1$ (3)

$0 \cdot C = 1$ (4)

Then, the solution of this system is $A = \frac{1}{4}$ , $B = -\frac{1}{2}$ and C = 0 .

And the equation of the plane which has intercepts of (4, 0, 0) and (0, -2, 0) and no z-intercept is $\frac{1}{4}\cdot x -\frac{1}{2}\cdot y = 1$ .

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Answer from: gymnastcaitlyn1

Given two points (4,0,0) and (0,-2,0) are two intercepts of plane Π which has no z-intercept.

First step: find direction vector of the two points:
V = <4-0, 0-(-2), 0-0> = <4,2,0>

Since there is no z-intercept, Π must be orthogonal to the z-axis, hence the unit normal vector is Z=<0,0,1>.

To find the normal vector of the plane Π, we take the cross-product of V & Z, i.e.
Π =
i j k
4 2 0
0 0 1
= <2*1, -4*1, 0>
= <2,-4,0>
The corresponding equation of the plane through (4,0,0) is then
2(x-4)-4y+0(z-0)=0 => 2x-4y=8
The equation of the plane is 2x-4y=8

Answer from: Quest

step-by-step explanation:

Answer from: Quest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

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Find the equation of the plane which has intercepts of (4,0,0) and (0,−2,0) and no z-intercept....

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