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

3D Geometry

11.1. Direction Ratio & Direction Cosines of a Line.

Q.1. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6. Write the direction cosines of a line parallel to the above line.

Solution :

We have, the line (4 – x)/2 = (y + 3)/3 = (z + 2)/6
Or, (x – 4)/–2 = (y + 3)/3 = (z + 2)/6
Comparing with (x – x1)/a = (y – y1)/b = (z – z1)/c
We get, a = – 2, b = 3, c = 6.
Let I, m, n be direction cosines of line parallel to given line.
Then l/a = m/b = n/c = 1/√(a² + b² + c²)
Or, l /–2 = m/3 = n/6 = 1/√(4 + 9 + 36) = 1/7
Therefore, l = – 2/7, m = 3/7, n = 6/7 [Ans.]

Q.2. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the direction cosines of a line parallel to this line.

Solution : – Do yourself. [Ans. = 2/7, 3/7, – 6/7]

11.2. Equation of a Line in Space.

Q.1. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P.

Solution :

Fig
We have the line x/1 = (y – 1)/2 = (z – 2)/3 = r ---------------------- (1)
Any point on the line (1) is Q (r, 2r + 1, 3r + 2).
The direction ratios of PQ are : r – 1, 2r + 1 – 6, 3r + 2 – 3
i.e. r – 1, 2r – 5, 3r – 1.
PQ is perpendicular to the line (1)
Therefore, 1(r – 1) + 2(2r – 5) + 3(3r – 1) = 0
=> 14r – 14 = 0 => r = 1.
Thus Q is (1, 3, 5), which is the required foot of perpendicular. [Ans.]
Also, PQ² = (1 – 1)² + (6 – 3)² + (3 – 5)² = 13
=> PQ = √13 units. [Ans.]

Q.2. Find the length and the foot of the perpendicular drawn from the point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11.

Solution :

Do yourself. [Ans. = Point (1, 2, 3); distance = √14 units.]

Paper By Mr. M.P.Keshari
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