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3. A line with positive direction cosines passes through the point P (2, -1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals
(A) 1
(B) √2
(C) √3
(D) 2
Sol. (C)
Direction Cosines of the line are 1/√3 , 1/√3 , 1/√3.
Any point on the line at a distance t from P(2, -1, 2) is
(2 + t/√3 , -1 + t/√3 , 2 + t/√3)
Which clearly lies on 2x + y + z = 9
Hence ,
=> t = √3
4. If the sum of the first n terms of an A.P. is cn2, then the sum of squares of these n terms is
(A) (n(4n2 - 1) c2)/6 (B) (n(4n2 + 1) c2)/3 (C) (n(4n2 - 1) c2)/3 (D) (n(4n2 + 1) c2)/6
tn = c {n2 - (n - 1)2}
= c (2n - 1)
=> tn2 = c2 (4n2 - 4n + 1)
=> ∑n=1n tn2 = c2 {4n(n + 1)(2n + 1)/6 - 4n(n + 1)/2 + n}
= c2n/3 {4(n+1)(2n+1) - 12(n+1)+6}
= c2n/3 {4n2 + 6n + 2 - 6n - 6 + 3} = c2/3 n(4n2 - 1)
MULTIPLE CORRECT CHOICE TYPE
5. The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose
(A) vertex is (2a/3 , 0)
(B) directrix is x = 0
(C) latus rectum is 2a/3
(D) focus is (a, 0)
Sol. (A, D)
G ≡ (h, k) => h = (2a + at2)/3, k = 2at/3 => ((3h-2a)/a) = (9k2)/4a2 => Hence , required parabola is 9y2)/4a2 =((3x - 2a))/a = 3/a ( x - 2a/3) => y2 = 4a/3 (x-2a/3) Vertex ≡ (2a/3,0); Focus ≡ (a, 0)
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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions...