CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

3D Geometry

Q.2. Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4).

Solution :

We have to find the equation of a plane through (0, – 1, – 1), (4, 5, 1) and
(3, 9, 4).
Let the plane through (0 – 1 – 1) be
a(x – 0) + b(y + 1) + c(z + 1) = 0 -------- (1)
This plane passes through the points (4, 5, 1) and (3, 9, 4).
Therefore, 4a + 6b + 2c = 0 and
3a + 10b + 5c = 0
Solving these two by cross multiplication, we get
a/(30 – 20) = b/(6 – 20) = c/(40 – 18)
Or, a/5 = b/ – 7 = c/11= k
Or, a = 5k, b = – 7k, c = 11k.
Putting these values of a, b and c, we get
5kx (– 7k)(y + 1) + 11k(z + 1) = 0
Or, 5x – 7y + 11z + 4 = 0. [Ans.]

Q.3. Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to each of the following planes :
2x + 3y – 3z = 2 and 5x – 4y + z = 6.

Solution :

Equation of the plane passing through (– 1, – 1, 2) is
a(x + 1) + b(x + 1) + c(x – 2 ) = 0 ------------------------ (1)
As (1) is perpendicular to
2x + 3y – 3z = 2 ------------------------ (2)
and 5x – 4y + z = 6 -------------------------- (3)
Therefore, 2a + 3b – 3c = 0 ------------------------- (4)
5a – 4b + c = 0 -------------------------- (5)
Eliminating a, b, c from (1), (4) and (5), we get
| x + 1 y + 1 z – 2 |
| 2 3 – 3 | = 0
| 5 –4 1 |
Or, (x + 1)(3 – 12) – (y + 1)(2 + 15) + (z – 2 )(– 8 – 15) = 0
Or, – 11(x + 1) – 17(y + 1) – 23(z – 2) = 0
Or, – 11x – 11 – 17y – 17 – 23z + 46 = 0
Or, – 11x – 17y – 23z + 18 = 0
Or, 11x + 17y + 23z – 18 = 0 [Ans.]

Q.4.Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z – 2)/5.

Solution :

Equation of the plane passing through (3, 4, 1) is
a(x – 3) + b(y – 4) + c(z – 1) = 0 --------------------- (1)
As, plane (1) passes through (0, 1, 0), hence
a(0 – 3) + b(1 – 4) + c(0 – 1) = 0
Or, – 3a – 3b – c = 0
Or, 3a + 3b + c = 0 ------------------------------- (2)
Also the plane (1) is parallel to the line (x + 3)/2 = (y – 3)/7 = (z – 2)/5
Therefore, 2a + 7b + 5c = 0 ------------------------------ (3)

Paper By Mr. M.P.Keshari
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