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

3D Geometry

Q.3. Find the point on the line : (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a distance 3√2 from the point (1, 2, 3).

Solution :

We have, (x + 2)/3 = (y + 1)/2 = (z – 3)/2 = λ (say)
Any point on this line is P(3λ – 2, 2λ – 1, 2λ + 3)
and the given point be Q(1, 2, 3).
PQ = 3√2. [Given]
Therefore, √{(3λ – 2 – 1)² + (2λ – 1 – 2)² + (2λ + 3 – 3)²} = 3√2
Or, √{(3λ – 3)² + (2λ – 3)² + (2λ)²} = 3√2
Squaring both sides, we get
9λ² + 9 – 18λ + 4λ² + 9 – 12λ + 4λ² = 18
Or, 17λ² – 30λ + 18 – 18 = 0
Or, 17λ² – 30λ = 0
Or, λ(17λ – 30) = 0
Therefore, λ = 0, λ = 30/17.
Hence, required point are P(0 – 2, 0 – 1, 0 + 3) or P(90/17 – 2, 60/17 – 1, 60/17 + 3)
Or, P(– 2, – 1 , 3) or P(56/17, 43/17, 111/17).

Q.4. Find the equation of the perpendicular drawn from the point (2, 4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 .

Solution :

.
Fig
The given equation of line is
(x + 5)/1 = (y + 3)/4 = (z – 6)/–9 = k (say) --------------- (1)
Any point on the line (1) is
M(k – 5, 4k – 3, – 9k + 6) ----------------------------------- (2)
Direction ratios of the lime PM is
k – 5 – 2, 4k – 3 – 4, – 9k + 6 + 1
= k – 7, 4k – 7, – 9k + 7
Direction ratio of the given line is proportional to 1, 4, – 9.
Therefore, line PM is perpendicular to the line (1)
Hence, 1.(k – 7) + 4.(4k – 7) + (– 9)(– 9k + 7) = 0
Or, k – 7 + 16k – 28 + 81k – 63 = 0
Or, k = 1.
Putting k = 1 in (2) we get the co-ordinates of M = (1 – 5, 4 – 3, – 9 + 6)
= (– 4, 1, – 3).
Equation of PM is given by
(x – 2)/(– 4 – 2) = (y – 4)/(1 – 4) = (z + 1)/(– 3 + 1)
Or, (x – 2)/– 6 (y – 4)/– 3 = (z + 1)/– 2
Or, (x – 2)/6 = (y – 4)/3 = (z + 1)/2 [Ans.]

Paper By Mr. M.P.Keshari
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