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

3D Geometry

Eliminating a, b, c from (1), (2) and (3) we get
| x – 3 y – 3 z – 1 |
| 3 3 1 | = 0
| 2 7 5 |
Or, (x – 3)(15 – 7) – (y – 4)(15 – 2) + (z – 1)(21 – 6) = 0
Or, 8(x – 3) – 13(y – 4) + 15(z – 1) = 0
Or, 8x – 24 – 13y + 52 + 15z – 15 = 0
Or, 8x – 13y + 15z + 13 = 0. [Ans.]

11.6. Distance of a Point from a Plane.

Q.1. Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.

Solution :

We have x + 2y + 4z = 38 ------------------------ (1)
Let image of the point P (1, 2, 3) in the line (1) be Q (α, β, γ).
Then PM is perpendicular to (1) and M is the mid-point of PQ.

Fig
Therefore, direction cosines of PM are perpendicular to 1, 2, 4 [which are d.r.s of the normal to (1)]
Thus equation of PM is
(x – 1)/1 = (y – 2)/2 = (z – 3)/4 ---------------- (2)
Any point on (2) is M (k + 1, 2k + 2, 4k + 3).
M lies on (1), hence we get
(k + 1) + 2(2k + 2) + 4(4k + 3) = 38 => k = 1.
Therefore, M (2, 4, 7).
Hence, 2 = (1 + α)/2 => α = 3; 4 = (2 + β)/2 => β = 6; 7 = (3 + γ)/2 => γ = 11.
Therefore, coordinates of Q are (3, 6, 11). [Ans.]

Q.2. Find the co-ordinates of the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0.

Solution :– Do yourself. [Ans. = (–3, 5, 2) ]

Q.3. From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and the co-ordinates of the foot of the perpendicular.

Solution :

Fig
The equation of the plane is
2x + y – 2z + 3 = 0 --------------------- (1)
Direction ratios of the plane are : 2, 1, – 2.
Therefore, direction ratios of line normal to the plane are : 2, 1, – 2.
Equation of the line PM is
(x – 1)/2 = (y – 2)/1 = (z – 4)/ –2 = r (say) [Ans.]
Therefore, coordinates of M = (2r + 1, r + 2, – 2r + 4)
Since the point M lies on (1)
Therefore, 2(2r + 1) + (r + 2) – 2 (– 2r + 4) + 3 = 0
Or, 4r + 2 + r + 2 + 4r – 8 + 3 = 0
Or, 9r – 1 = 0
Or, r = 1/9.
Therefore, foot of perpendicular = (2/9 + 1, 1/9 + 2, – 2/9 + 4)
= (11/9, 19/9, 34/9) [Ans.]
Length of perpendicular from (1, 2, 4) is
PM = |{2(1) + 2 – 2(4) + 3}/√(4 + 1 + 4)| = 1/3 unit. [Ans.]

Paper By Mr. M.P.Keshari
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