Find the intersection of the plane a: 3x-4y Z 7 = 0 and the line L: x-3y + 12 = 0, 2y-z-6 = 0 The equation of a line perpendicular to a line in a given plane

Find the intersection of the plane a: 3x-4y Z 7 = 0 and the line L: x-3y + 12 = 0, 2y-z-6 = 0 The equation of a line perpendicular to a line in a given plane

3x-4y Z 7 = 0 default + otherwise it can't be done
3x-4y +z +7=0
x-3y+12=0
2y-z-6 = 0, the intersection is (3,5,4)
The line l can be reduced to (x-3) / 3 = (Y-5) / 1 = (Z-4) / 2
The direction vector is n = (3,1,2)
Let a (x0, Y0, Z0) be a point on the line P, and pass (3,5,4), then the linear equation is
(x-3) / (3-x0) = (Y-5) / (5-y0) = (Z-4) / (4-z0) (1) the direction vector is m = (3-x0,5-y0,4-z0)
M = 0, that is, 3 (3-x0) + (5-y0) + 2 (4-z0) = 0 (2)
A in plane a 3x0-4y0 + Z0 + 7 = 0 (3)
The normal vector of plane a = (3, - 4,1) a.m = 0, that is, 3 (3-x0) + 4 (5-y0) + (4-z0) = 0 (4)
From (2) (3) (4), x0 = Y0 = Z0 = can be substituted into (1)

Given the equations 2x-y + Z = 5,5x + 6y-z = 9, what is the value of X + y?

x=0
y=2
z=3

Given that | x | = 1, | y | = 2, | Z | = 3, and XY ＜ 0, XYZ ＞ 0, try to find the value of (x + y + Z) × (XY + YZ)

xy

The triple integral ∫ ∫ ∫ zdxdydz is calculated, where Ω is a closed domain bounded by z = x ^ 2 + y ^ 2 and z = 4
2π)dθ∫(0~2)ρdρ∫(ρ^2~4)zdz
Why is the lower bound of the integral for Z ρ ^ 2?

Because the equation of surface z = x ^ 2 + y ^ 2 in cylindrical coordinates is Z = ρ ^ 2
If the problem is to calculate the integral value, the positive solution is as follows:
Because the area of the area between the plane of Z = constant and Ω is π Z
So ∫ zdxdydz = ∫ (0 ~ 4) Z (π z) DZ = (1 / 3) π (Z ^ 3) (0 ~ 4) = 64 π / 3

Calculate the triple integral ∫∫∫∫ zdxdydz, Ω is a closed region bounded by x ^ 2 + y ^ 2 + Z ^ 2 = 4 and z = 1 / 3 (x ^ 2 + y ^ 2)
Select appropriate coordinate system to calculate

Both are cylindrical coordinate methods

Calculate the triple integral ∫ ∫ zdxdydz, where Ω is the region bounded by z = x ^ 2 + y ^ 2, z = 0, x ^ 2 + y ^ 2 = 1
The key problem is how to determine the scope of XYZ

It is convenient to use cylindrical coordinates
Integral limit: 0 ≤ θ ≤ 2 π, 0 ≤ R ≤ 1, 0 ≤ Z ≤ R & # 178;, dxdydz = rdrd θ DZ
∫ dθ∫rdr∫zdz
=∫ dθ∫(1/2)r^5dr∫
=(1/12)∫dθ
=π/6

The triple integral ∫∫∫∫ xydxdydz is calculated, where Ω is the closed region bounded by three coordinate planes and plane x + y + Z = 1

Just use rectangular coordinates

The triple integral ∫∫∫ ZDV is calculated, where Ω is the region bounded by the upper spherical z = root (4-x ^ 2-y ^ 2) and the plane z = 0
be deeply grateful!

This is the area surrounded by cylinder, cone and z = 0. You need to be able to draw by yourself
The simplest way to solve this problem is the section method (2 first and 1 later)
The cross section of the solid is a circle DZ: 1 ≤ X & # 178; + Y & # 178; ≤ 4-z & # 178;
When X & # 178; + Y & # 178; = 1, Z in the cone = √ 3, so the range of Z is: 0 →√ 3
In this paper, we first do double integration on DZ, and then do definite integration on Z
∫∫∫zdv
=∫ [0 → √ 3] ZDZ ∫ (DZ) DXDY, where DZ: 1 ≤ X & # 178; + Y & # 178; ≤ 4-z & # 178;
This double integral is very simple. Because the integrand is 1, the integral result is the area of the ring π (4-z & # 178; - 1) = π (3-z & # 178;)
=π∫[0→√3] z(3-z²) dz
=π∫[0→√3] (3z-z³) dz
=π[(3/2)z²-(1/4)z^4] |[0→√3]
=π(9/2-9/4)
=9π/4