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Solution:
Let us consider
According to the equations line P1 passes through the point P(2, 5, 0)
And the equation of a plane containing line P2 is
a(x - 0) + b(y + 1) + c(z - 1) = 0 -(1)
Where 2a - b + 2c = 0
If it is parallel to line P1 then
-a + 2b + 3c = 0
So,
Now, substitute the value of a, b, c in the eq(1) we get
a(x - 0) + b(y + 1) + c(z - 1) = 0
-7(x - 0) - 8(y + 1) + 3(z - 1) = 0
-7x - 8y - 8 + 3z - 3 = 0
7x + 8y - 3z + 11 = 0 -(2)
So, this is the equation of the plane that contain line P2 and parallel to line P1.
Hence, the shortest distance between P1 and P2 = Distance between point P(2, 5, 0) and plane (2)
Solution:
Let us consider
Let us assume the equation of the plane containing P1 is a(x + 1) + b(y + 1) + c(z+1) = 0
Plane is parallel to P1 = 7a - 6b + c = 0 -(1)
Plane is parallel to P2 = a - 2b + c = 0 -(2)
On solving eq(1) and eq(2), we get,
The equation of the plane is -4(x + 1) - 6(y + 1) - 8(z + 1) = 0
Final equation of plane is 4(x + 1) + 6(y + 1) + 8(z + 1) = 0
Solution:
The equation of a plane containing the line 3x - y - 2z + 4 = 0, 2x + y + z + 1 = 0 is
x(2λ + 3) + y(λ - 1) + z(λ - 2) + λ + 4 = 0 -(1)
If it is parallel to the line then,
2(2λ + 3) + 4(λ - 1) + (λ - 2) = 0
λ = 0
On putting λ = 0 in eq(1) we get,
3x - y - 2z + 4 = 0 -(2)
As this equation of the plane consist the second line and parallel to the first line.
It is clear that the line passes through the point (1, 3, -2)
So, the shortest distance 'D' between the given lines is equal to the
length of perpendicular from point (1, 3, -2) on the plane (2)
D =
Exercise 29.14 in RD Sharma Class 12 typically deals with finding the equation of a plane under various conditions. Key concepts include: