Let A (2, –3) and B(–2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line
2x + 3y = 9
2x – 3y = 7
3x + 2y = 5
3x – 2y = 3
A.
2x + 3y = 9
If C be (h, k) then centroid is (h/3, (k – 2)/3) it lies on 2x + 3y = 1.
⇒ locus is 2x + 3y = 9.
The curve y = (λ + 1)x^{2} + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals -
{–2, 2}
{1}
{-2}
{2}
C.
{-2}
for point of intersection at exactly one point
λx + 3 = (λ + 1)x^{2} + 2
(λ + 1)x^{2} – λx – 1 = 0
Δ = 0
λ^{2} + 4(λ + 1) = 0
λ^{2} + 4λ + 4 = 0
(λ + 2)^{2} = 0 λ = – 2
The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that
it passes through the origin
it makes angle π/2 + θ with the x-axis
it passes through (aπ/2 ,-a)
it is at a constant distance from the origin
D.
it is at a constant distance from the origin
Clearly dy/dx = an θ
⇒ slope of normal = - cot θ
Equation of normal at ‘θ’ is
y – a(sin θ - θ cos θ) = - cot θ(x – a(cos θ + θ sin θ)
⇒ y sin θ - a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.
If the straight line y = mx + c (m > 0) touches the parabola y^{2} = 8(x + 2), then the minimum value taken by c is
12
8
4
6
C.
4
The tangent of slope m must be of the form y = m(x + 2) + a/m
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
3/2
5/2
7/2
9/2
C.
7/2
The distance between 4x + 2y + 4z - 16 = 0 and 4x + 2y + 4z + 5 = 0 is
If the straight lines x = 1 + s, y = –3 – λs, z = 1 + λs and x = t/ 2 , y = 1 + t, z = 2 – t with parameters s and t respectively, are co-planar then λ equals
–2
–1
-1/2
0
A.
–2
Given are coplanar then plan passing through these lines has normal perpendicular to these lines
⇒ a - bλ + cλ = 0 and a/2 +b -c =0 (where a, b, c are direction ratios of the normal to the plan) On solving, we get λ = -2.
A line makes the same angle θ, with each of the x and z-axis. If the angle β, which it makes with y-axis, is such that sin^{2}β = 3sin^{2}θ , then cos2θ equals
2/3
1/5
3/5
2/3
C.
3/5
A line makes angle θ with x-axis and z-axis and β with y-axis.
∴ l = cosθ, m = cosβ,n = cosθ
We know that, l^{2}+ m^{2}+ n^{2}= 1
cos^{2}θ + cos^{2}β +cos^{2}θ =1
2cos^{2}θ = 1- cos^{2}β
2cos^{2}θ = sin^{2}β
But sin^{2}β = 3 sin^{2}θ
therefore from equation (i) and (ii)
3sin^{2}θ = 2cos^{2}θ
3(1-cos^{2}θ) = 2cos^{2}θ
3-3cos^{2}θ = 2cos^{2}θ
3 = 5cos^{2}θ
The intersection of the spheres x^{2} +y^{2} +z^{2} + 7x -2y-z =13 and x^{2} +y^{2} +z^{2} -3x +3y +4z = 8 is the same as the intersection of one of the sphere and the plane
x-y-z =1
x-2y-z =1
x-y-2z=1
2x-y-z =1
D.
2x-y-z =1
Required plane is S_{1} – S_{2} = 0 where
S_{1} = x^{2} + y^{2} + z^{2} + 7x – 2y – z – 13 = 0
and S_{2} = x^{2} + y^{2} + z^{2} – 3x + 3y + 4z – 8 = 0
⇒ 2x – y – z = 1.
The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:
x – 2y + 11 = 0
x + 2y + 11 = 0
x + 2y – 11 = 0
x – 2y – 11 = 0
B.
x + 2y + 11 = 0
Equation of the required plane is
(x + y + z – 6) + λ(2x + 3y + z + 5) = 0
i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0
This plane is perpendicular to xy plane whose
equation is z = 0
i.e. 0 . x + 0 . y + z = 0
∴ By condition of perpendicularity
0 .(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0
i.e. λ = –1
∴ Equation of required plane is
(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) =0
or x + 2y + 11 = 0
A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by
(3a, 3a, 3a), (a, a, a)
(3a, 2a, 3a), (a, a, a)
(3a, 2a, 3a), (a, a, 2a)
(2a, 3a, 3a), (2a, a, a)
B.
(3a, 2a, 3a), (a, a, a)
Any point on the line and any point on the line
Now direction cosine of the lines intersecting the above lines is proportional to (2t_{2} – a – t_{1}, t_{2} – t_{1} + a, t_{2} – t_{1}).
Hence 2t_{2} – a – t_{1} = 2k , t_{2} – t_{1} + a = k and t_{2} – t_{1} = 2k
On solving these, we get t_{1} = 3a , t_{2} = a. Hence points are (3a, 2a, 3a) and (a, a, a).