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

3D Geometry

Q.4. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).

Solution :

The equation of plane through A(3, – 1, 2) is
a(x – 3) + b(y + 1) + c(z – 2) = 0 -------------- (1)
This passes through B(5, 2, 4), then
a(5 – 3) + b(2 + 1) + c(4 – 2) = 0
Or, 2a + 3b + 2c = 0 ----------------------- (2)
(1) also passes through (– 1, – 1, 6) , then
a(– 1 – 3) + b(– 1 + 1) + c(6 – 2) = 0
Or, – 4a + 4c = 0 -------------------------- (3)
Solving (2) and (3) by cross multiplication, we get
a/( 3 – 0) = b/(– 2 – 2) = c/(0 + 3) = k
a/3 = b/– 4 = c/3 = k
a = 3k, b = – 4k, c = 3k.
Putting the value of a, b and c in (1), we get
3(x – 3) – 4(y + 1) + 3(z – 2) = 0
Or, 3x – 9 – 4y – 4 + 3z – 6 = 0
Or, 3x – 4y + 3z – 19 = 0 ---------------------- (4)
Distance of the point P(6, 5, 9) from the plane (4)
= |[3(6) – 4(5) + 3(9) – 19]/√(9 + 16 + 9)|
= |[18 – 20 + 27 – 19]/√(34)|
= 6/√(34). [Ans.]

Q.5. Find the co-ordinates of the point where the line (x + 1)/2 = (y + 2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6.

Solution :

Fig.
We have the equation of lines
(x + 1)/2 = (y + 2)/3 = (z + 3)/4 = r (say).
Any point on the line is given by P(2r – 1, 3r – 2, 4r – 3)
As the point P lies on the plane x + y + 4z = 6
Therefore, (2r – 1) + (3r – 2) + 4(4r – 3) = 6
Or, 2r – 1 + 3r – 2 + 16r – 12 = 6
Or, 21r – 15 = 6
Or, 21r = 21 => r = 1.
Hence, required point is P(2 – 1, 3 – 2, 4 – 3) = P(1, 1, 1) [Ans.]

Q.6. Find the distance of the point (– 2, 3 – 4) from the line
(x + 2)/3 = (2y + 3)/4 = (3z + 4)/5
measured parallel to the plane 4x + 12y – 3z + 1 = 0.

Solution :

Let the given point be P (– 2, 3, – 4)
Any point on the given line
(x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 = k
is Q{3k – 2, (4k – 3)/2 , (5k – 4)/3.
Direction ratios of the line PQ is 3k – 2 + 2, (4k – 3)2 – 3, (5k – 4)/3 + 4
= 3k, (4k – 3 – 6)/2, (5k – 4 + 12)/3
= 3k, (4k – 9)/2, (5k + 8)/3.
As the line PQ is parallel to the plane
4x + 12y – 3z + 1 = 0.
Therefore, 4.(3k) + 12.(4k – 9)/2 – 3.(5k + 8)/3 = 0 => k = 2.
Therefore, co-ordinate of Q = {6 – 2, (8 – 3)/2, (10 – 4)/3} = (4, 5/2, 2).
Therefore, length PQ = √{(4 + 2)² + (5/2 – 3)² + (2 + 4)²}
= √(36 + 1/4 + 36)
= √(72 + 1/4)
= √(289/4)
= 17/2.[Ans.]

Paper By Mr. M.P.Keshari
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