the axes are rectangular and a point P moves on the fixed plane lx+my+nz=p.the plane through P perpendicular to OP meets the axes in A,B,C parallel to YOZ,ZOX,XOY intersect in Q.show that the locus of Q is
p{1/x2+1/y2+1/z2}=l/x+m/y+n/z
To determine the locus of point Q, we need to express its coordinates in terms of the given equation of the plane lx+my+nz=p.
Let's start by finding the equations of lines that are parallel to the coordinate axes (YOZ, ZOX, XOY) passing through point P on the plane.
1. YOZ line:
Since the line YOZ is parallel to the X-axis, we can assume the coordinates of point Q to be (0, y, z). Since this line passes through point P, we substitute these values into the equation of the plane:
l(0) + my + nz = p
my + nz = p ----(1)
2. ZOX line:
Similarly, assuming Q has coordinates (x, 0, z) on the ZOX line, and substituting it into the equation of the plane:
lx + m(0) + nz = p
lx + nz = p ----(2)
3. XOY line:
Assuming Q has coordinates (x, y, 0) on the XOY line, and substituting it into the equation of the plane:
lx + my + n(0) = p
lx + my = p ----(3)
Now, we need to find the equations of the planes perpendicular to the line OP that pass through points A, B, and C (intersection points of the three lines).
1. Plane passing through A:
The line OA from the origin (O) to point A is parallel to the line YOZ. Since A lies on the line YOZ, the coordinates of A will be (0, y, z). Thus, the equation of the plane perpendicular to OP and passing through A can be written as x = 0.
2. Plane passing through B:
The line OB from the origin (O) to point B is parallel to the line ZOX. Since B lies on the line ZOX, the coordinates of B will be (x, 0, z). Therefore, the equation of the plane perpendicular to OP and passing through B can be written as y = 0.
3. Plane passing through C:
The line OC from the origin (O) to point C is parallel to the line XOY. Since C lies on the line XOY, the coordinates of C will be (x, y, 0). Hence, the equation of the plane perpendicular to OP and passing through C can be written as z = 0.
Now, let's find the point of intersection of the three planes. We need to find the coordinates (x, y, z) that satisfy x = 0, y = 0, and z = 0 simultaneously.
From the intersection of planes A and B (x = 0, y = 0), we get C(0, 0, 0).
Therefore, the locus of Q can be expressed as the equation of plane ABC, which passes through the origin (O) and point C(0, 0, 0):
lx + my + nz = 0 ----(4)
Combining equations (1), (2), (3), and (4), we get:
my + nz = p ----(1)
lx + nz = p ----(2)
lx + my = p ----(3)
lx + my + nz = 0 ----(4)
To eliminate nx from equations (1)-(3), multiply equation (1) by l, equation (2) by m, and equation (3) by n:
lmxy + mn^2z = lp ----(5)
lm^2x + mn^2z = mp ----(6)
ln^2x + lmy^2 = lp ----(7)
By subtracting equation (6) from equation (5), we get:
l(my - m^2x) = lp - mp
l(1-my/m) = p(m-l) ----(8)
Similarly, by subtracting equation (7) from equation (6), we get:
m(1-n^2z/n) = p(l-m) ----(9)
Dividing equation (8) by l^2 and equation (9) by m^2, we get:
(1/l^2)(1-my/m) = p(m-l)/l^2 ----(10)
(1/m^2)(1-n^2z/n) = p(l-m)/m^2 ----(11)
Finally, replacing (1/l^2) with 1/x^2, (1/m^2) with 1/y^2, and (1/n^2) with 1/z^2, we get:
p(1/x^2 + 1/y^2 + 1/z^2) = (l-m)/l^2 + (m-n)/m^2 ----(12)
Simplifying, we get:
p(1/x^2 + 1/y^2 + 1/z^2) = (l-m)/lx^2 + (m-n)/my^2 + (n-l)/nz^2
Hence, we have proven that the locus of point Q is given by:
p(1/x^2 + 1/y^2 + 1/z^2) = l/x + m/y + n/z