If a variable plane in 3-D space moves in such a way that the sum of its reciprocal of intercepts on the x and y-axis exceeds, the reciprocal of its intercept on the z-axis by 2, then all such planes will pass through the point:

Question

If a variable plane in 3-D space moves in such a way that the sum of its reciprocal of intercepts on the x and y-axis exceeds, the reciprocal of its intercept on the z-axis by 2, then all such planes will pass through the point:

Answer 1

The plane with intercepts a,b,c is

x/a + y/b + z/c = 1
Now, you also have 1/a + 1/b = 1/c + 2
So express c in terms of a and b, and find the sum k = 1/a+1/b+1/c.

Now, it is easy to find proofs online that if 1/a + 1/b + 1/c = k, then the plane passes through (1/k, 1/k, 1/k)

Answer 2

That's what I get.

Answer 3

Thank u

So answer will be (1/2,1/2,-1/2)

Answer 4

To solve this problem, we need to understand the concept of intercepts and how they relate to the equation of a plane in 3D space.

In 3D space, the equation of a plane can be written in intercept form as:

x/a + y/b + z/c = 1

where a, b, and c are the intercepts of the plane on the x, y, and z-axis respectively.

Let's analyze the information given in the problem statement. It states that the sum of the reciprocal of the x and y intercepts exceeds the reciprocal of the z intercept by 2. Mathematically, we can write this as:

1/a + 1/b = 1/c + 2

We want to find the point through which all such planes pass. In other words, we need to find the common point of intersection of all these planes.

To find this point, we can choose any two planes satisfying the given condition and find their common point of intersection. Since there are infinitely many such planes, we can assume arbitrary values for a, b, and c and solve for the common point.

Let's assume a = 1, b = 1, and c = 1. Plugging these values into the equation of the plane, we get:

x/1 + y/1 + z/1 = 1
x + y + z = 1

Now, let's assume a = 2, b = 2, and c = 2. Plugging these values into the equation of the plane, we get:

x/2 + y/2 + z/2 = 1
x + y + z = 2

By solving these two equations simultaneously, we can find the common point of intersection. Subtracting the first equation from the second equation, we get:

2 - 1 = 1

So, the common point of intersection is (1, 1, 1).

Therefore, all such planes pass through the point (1, 1, 1).