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
Give answer
To find the locus of point Q, we need to express the coordinates of Q in terms of p, l, m, and n. Let's go step by step to derive the equation.
Step 1: Find the coordinates of point P.
The equation of the plane given is: lx + my + nz = p. If we denote the coordinates of P as (x, y, z), then the equation can also be written as: lx + my + nz - p = 0. Therefore, the coordinates of point P are (x, y, z).
Step 2: Find the equation of the plane through P perpendicular to OP.
The vector OP is given by the coordinates of P, so OP = (x, y, z). The equation of the perpendicular plane can be written as Ax + By + Cz + D = 0, where A, B, C, and D are constants to be determined.
To find A, B, C, and D, we know that the plane passes through point P, so substituting the coordinates of P into the equation, we get:
A * x + B * y + C * z + D = 0.
We also know that the plane is perpendicular to vector OP, which means the dot product of the normal vector of the plane and vector OP is zero. The normal vector of the plane is (A, B, C), and the dot product with OP is: A * x + B * y + C * z. Equating it to zero, we have:
A * x + B * y + C * z = 0.
Comparing this equation with lx + my + nz - p = 0, we can conclude that:
A = l, B = m, C = n, and D = -p.
Therefore, the equation of the plane through P perpendicular to OP is:
l * x + m * y + n * z - p = 0.
Step 3: Find the coordinates of point Q.
To find the coordinates of point Q, we need to find the intersection of the three planes. The equations of the three planes are:
Plane through Q parallel to YOZ: x = l/Qy + m/Qz - n/Q;
Plane through Q parallel to ZOX: y = -l/Qx + m/Qz + n/Q;
Plane through Q parallel to XOY: z = l/Qx - m/Qy + n/Q.
To find the coordinates of Q, we need to solve these three equations simultaneously.
Step 4: Substitute the coordinates of Q into the equation.
Substituting the coordinates of Q (x, y, z) into the equation lx + my + nz - p = 0, we get:
l * (l/Qy + m/Qz - n/Q) + m * (-l/Qx + m/Qz + n/Q) + n * (l/Qx - m/Qy + n/Q) - p = 0.
Simplifying this equation, we get:
l^2/Qy + m^2/Qz + n^2/Qx - l^2/Qx + m^2/Qz + n^2/Qy + l^2/Qx - m^2/Qy + n^2/Qz - p = 0.
Combining like terms, we have:
(1/Q)(l^2/x + m^2/y + n^2/z) - p = 0.
Therefore, the equation of the locus of Q is:
1/x^2 + 1/y^2 + 1/z^2 = (p/Q)(1/l^2 + 1/m^2 + 1/n^2).
Finally, simplifying the expression (p/Q)(1/l^2 + 1/m^2 + 1/n^2), we get:
(p/Q)(1/x^2 + 1/y^2 + 1/z^2) = 1/l^2 + 1/m^2 + 1/n^2.
Hence, the locus of point Q can be represented by the equation:
(p/Q)(1/x^2 + 1/y^2 + 1/z^2) = 1/l^2 + 1/m^2 + 1/n^2.
To solve this problem, we need to follow these steps:
Step 1: Determine the equation of the plane through point P perpendicular to OP.
Step 2: Find the coordinates of points A, B, and C where the plane intersects the axes.
Step 3: Find the equations of lines formed by points A, B, and C.
Step 4: Determine the coordinates of point Q, where the lines intersect.
Step 5: Create an equation based on the coordinates of point Q.
Let's begin with step 1:
Step 1: Equation of the plane through point P perpendicular to OP
The equation of a plane can be written in the form ax + by + cz = d, where (a, b, c) is the normal vector to the plane, (x, y, z) represents any point on the plane, and d is a constant.
In this case, we have the plane lx + my + nz = p passing through point P. Since OP is perpendicular to the plane, the vector (l, m, n) is orthogonal to the vector (x, y, z) - (0, 0, 0). Therefore, (l, m, n) is the normal vector to the plane.
Step 2: Coordinates of points A, B, and C
When the plane intersects the axes, the coordinates of the points of intersection can be found by substituting the appropriate axis values into the equation of the plane.
For the x-axis, y = 0 and z = 0, so substituting these values into the equation lx + my + nz = p, we get A(p/l, 0, 0).
Similarly, for the y-axis, x = 0 and z = 0, so substituting these values into the equation lx + my + nz = p, we get B(0, p/m, 0).
For the z-axis, x = 0 and y = 0, so substituting these values into the equation lx + my + nz = p, we get C(0, 0, p/n).
Step 3: Equations of lines formed by points A, B, and C
The equations of the lines formed by points A, B, and C can be written in vector parametric form.
For the line passing through A and parallel to the yoz-plane, we can write the equation as r = A + t(a, b, c), where t is a variable parameter. Simplifying this equation, we get x = p/l, y = t, and z = t.
Similarly, for the line passing through B and parallel to the zox-plane, we get x = t, y = p/m, and z = t.
For the line passing through C and parallel to the xoy-plane, we get x = t, y = t, and z = p/n.
Step 4: Coordinates of point Q
To find the coordinates of point Q, we need to find the intersection of the three lines formed by points A, B, and C.
Substituting the equations of lines formed by A, B, and C, we equate the x, y, and z values to find t.
From the equations x = p/l and y = t, we can equate them to find t = y/p. Similarly, from the equations y = p/m and z = t, we find t = z/p. From the equations z = p/n and x = t, we find t = x/p.
Since all the values of t are equal, we can write t = x/p = y/p = z/p.
Step 5: Equation of the locus of Q
We can substitute the value of t into any of the equations x = p/l, y = p/m, or z = p/n to get an equation containing only x, y, and z.
Let's choose the equation z = p/n. Substituting t = z/p, we get z = (p/n)z/p. Simplifying this equation, we obtain 1/z = n/p.
Similarly, substituting t = x/p into the equation y = p/m, we get 1/y = m/p, and substituting t = y/p into the equation x = p/l, we get 1/x = l/p.
Putting all these together, we arrive at the final equation:
1/x^2 + 1/y^2 + 1/z^2 = (l/p)^2 + (m/p)^2 + (n/p)^2.
Therefore, the locus of point Q can be represented by the equation:
p(1/x^2 + 1/y^2 + 1/z^2) = l/x + m/y + n/z.
This concludes the solution to the problem.