问题1.写出在坐标轴上截距为2，-3和4的平面方程。
解决方案：
Given: Intercept on the coordinate axes are 2,-3 and 4
We represent equation of a plane whose intercepts on
the coordinate axes are p, q and r respectively as follows,
(x / p) + (y / q) + (z / r) = 1 -Equation (1)
Here, p = 2, q = -3, r = 4,
The required equation of plane is
x / 2 + y / -3 + z / 4 = 1
6x – 4y + 3z / 12 = 1
6x – 4y + 3z = 12
问题2(i)。简化以下平面的方程以截取形式并在坐标轴上找到截距：4x + 3y – 6z -12 = 0。
解决方案：
Reduce the equation 4x + 3y – 6z -12 = 0 to intercept form
4x + 3y – 6z -12 = 0 – Equation (1)
Divide equation(1) by 12
4x / 12 + 3y / 12 – 6z / 12 = 12 / 12
x / 3 + y / 4 + z / -2 = 1 -Equation (2)
This is in the form x / a + y / b + z / c = 1 -Equation (3)
On comparing equation (2) and equation (3), we get
a = 3, b = 4, c = -2
So, the intercepts on coordinate axes are 3, 4, -2.
问题2(ii)。简化以下平面的方程以截取形式并在坐标轴上找到截距：2x + 3y – z = 6。
解决方案：
Reduce 2x + 3y – z = 6 in intercept form :
2x + 3y – z = 6 -Equation(1)
Divide equation (1) by 6,
2x / 6 + 3y / 6 – z / 6 = 6 / 6
x / 3 + y / 2 + z / -6 = 1 -Equation (2)
Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,
x / a + y / b + z / c = 1 -Equation (3)
On comparing equation (2) and equation (3)
We get a = 3, b = 2, c = -6
So, The intercepts on coordinate axes by given plane are 3, 2, -6
问题2(iii)。简化以下平面的方程以截取形式并在坐标轴上找到截距：2x – y + z = 5。
解决方案：
To find intercepts on coordinate axes by plane 2x – y + z = 5
2x – y + z = 5 -Equation(1)
Divide equation (1) by 5,
2x / 5 – y / 5 + z / 5 = 5 / 5
x / ( 5 /2 ) + y /-5 + z / 5 = 1 -Equation (2)
Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,
x / a + y / b + z / c = 1 -Equation (3)
On comparing equation (2) and equation (3)
We get a = 5 / 2, b = -5, c = 5
So, the intercepts on coordinate axes by given plane as 5 / 2, -5, 5.
问题3.假定三角形ABC的质心为点(α，β，γ)，找到与轴A，B和C相交的平面方程。
解决方案：
Given that plane meets axes in A, B, and C
Let, A = (a, 0, 0), B = (0, b, 0), c = (0, 0, c)
We have centroid of ABC as (α, β, γ). Centroid of plane ABC is given by,
Centroid = (x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3, (z1 + z2 + z3) / 3
(α, β, γ) = [(a + 0 + 0) /3, (0 + b + 0) / 3, (0 + 0 + c) / 3]
(α, β, γ) = [a / 3, b / 3, c / 3]
We get,
a / 3 = α ⇒ a = 3α -Equation(1)
b / 3 = β ⇒ b = 3β -Equation(2)
c / 3 = γ ⇒ c = 3γ -Equation(3)
Equation of intercept form of plane with a, b, c
as intercepts on coordinate axes is,
x / a + y / b + z / c = 1
Put a, b, c from equation (1),(2) and (3),
x / 3α + y / 3β + z / 3γ = 1
Multiplying by 3 on both sides,
3x / 3α + 3y / 3β + 3z / 3γ = 3
x / α + y / β + z / γ = 3
问题4：找到通过点(2、4、6)并在坐标轴上进行相等截距的平面方程。
解决方案：
Intercepts on coordinate axes are equal,
Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,
x / a + y / b + z / c = 1
From given condition let, a = b = c = p (say)
x / p + y / p + z / p = 1
x + y + z / p = 1
x + y + z = p -Equation(1)
Plane is passing through the point (2, 4, 6). By using equation(1)
x + y + z = p
2 + 4 + 6 = p
12 = p
Substitute p in equation(1)
x + y + z = 12
So the required equation of plane is given by,
x + y + z = 12
问题5.平面分别在A，B和C处与坐标轴相交，因此三角形ABC的质心为(1 、、-2、3)。找到平面方程。
解决方案：
Given that plane meets the coordinate axes at A, B, C.Centroid of plane ABC is (1, -2, 3)
Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,
x / a + y / b + z / c = 1 -Equation(1)
Centroid =(x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3, (z1 + z2 + z3) / 3
(1, -2, 3) = [(a + 0 + 0) /3, (0 + b + 0) / 3, (0 + 0 + c) / 3]
(1, -2, 3) = [a / 3, b / 3, c / 3]
Now by comparing we get,
a / 3 = 1 ⇒ a = 3 -Equation(2)
b / 3 = -2 ⇒ b = -6 -Equation(3)
c / 3 = 3 ⇒ c = 9 -Equation(4)
Substitute a, b, c in equation (1) to get the equation of required plane
x / 3 + y / -6 + z / 9 = 1
6x – 3x + 2z / 18 = 1
6x – 3x + 2z = 18