Respuesta :
The distance between the xy, yz and xz plane are 5, 3 and 7.
Multivariable Calculus
This question relates to multivariable calculus.
a) Distance of point (x, y, z) from the xy-plane
The |z co-ordinate of the point| = |z|
distance of point (3, 7, -5) from xy plane
[tex]|-5|=5[/tex]
b) Distance of point (x, y, z) from yz plane
The |x co-ordinate of the point| = |x|
Therefore, the distance of point (3, 7, -5) from yz plane is
[tex]|3|=3[/tex]
c) Distance of point (x, y, z) from xz plane
The |y co-ordinate of the point| = |y|.
The distance of the point (3, 7 -5) from the xz-plane is
[tex]|7|=7[/tex]
d) Distance of point (x, y, z) from the x-axis
The distance of the points from x-axis is
[tex]x^2 = y^2+z^2\\x=\sqrt{y^2+z^2}\\x= \sqrt{7^2+(-5)^2}\\ x=\sqrt{74}[/tex]
e) Distance of point (x, y, z) from the y-axis
[tex]y=\sqrt{x^2+z^2}\\y=\sqrt{3^2+(-5)^2}\\y=\sqrt{34} \\[/tex]
f) Distance of point (x, y, z) from the z-axis
[tex]z=\sqrt{x^2+y^2}\\ z=\sqrt{3^2+7^2}\\ z=\sqrt{58}[/tex]
From the calculations above,
- The distance of (3, 7, -5) from the xy plane = 5
- The distance of (3, 7, -5) from the yz-plane = 3
- The distance of (3, 7, -5) from the xz-plane = 7
- The distance of (3, 7, -5) from the x-axis = [tex]\sqrt{74}[/tex]
- The distance of (3, 7, -5) from the y-axis = [tex]\sqrt{34}[/tex]
- The distance of (3, 7, -5) from the z-axis = [tex]\sqrt{58}[/tex]
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