Q1.A(3,2,0),B(5,3,2) and C(-9,6,-3) are the vertices of a triangle ABC. If the bisector of ∠ABC meets BC at D, then coordinates of D are
  •  (-19/8, 57/16, 17/16)
  •  (19/8, -57/16, 17/16)
  •  None of these
  •  (19/8, 57/16, 17/16)
D divides BC in the ratio =AB∶AC i.e. 3∶13 Therefore, coordinates of D are ((3×-9+13×5)/(3+13),(3×6+13×3)/(3+13),┤ ├ (3×-3+13×2)/(3+13)) or,(19/8,57/16,17/16)

Q2.The plane x/2+y/3+z/4=1, cuts the axes in A,B,C, then the area of the ∆ABC, is
  •  √29 sq units
  •  √61 sq units
  •  √41 sq units
  •  None of these
The given equation of plane is x/2+y/3+z/4=1 On comparing with x/a+y/b+z/c=1, we get a=2,b=3,c=4 Area of ∆ ABC=1/2 √(a^2 b^2+b^2 c^2+c^2 a^2 ) ∆=1/2 √(4×9+9×16+16×4) =1/2 √(36+144+64)=1/2 √244=√61

Q3.  The direction cosines of the line 6x-2=3y+1=2z-2 are
  •   1/√3,1/√3,1/√3
  •  1, 2, 3
  •  None of these
  •  1/√14,2/√14,3/√14

Q4. The equation of the sphere concentric with the sphere 2x^2+2y^2+2z^2-6x+2y-4z=1 and double its radius is
  •  x^2+y^2+z^2-x+y-z=1
  •  x^2+y^2+z^2-6x+2y-4z=1
  •  2x^2+2y^2+2z^2-6x+2y-4z-15=0
  •  2x^2+2y^2+2z^2-6x+2y-4z-25=0
Given equation of sphere is x^2+y^2+z^2-3x+y-2z-1/2=0 where centre is(3/2,-1/2,1) and radius of sphere is√(9/4+1/4+1+1/2)=2 equation of family of concentric sphere is x^2+y^2+z^2-3x+y-2z+λ=0 ….(i) ∴ According to question, √(9/4+1/4+1-λ)=4 ⟹14/4-λ=16 ⟹ λ=-25/2 ∴ From Eq. (i), x^2+y^2+z^2-3x+y-2z-25/2=0 ⟹2x^2+2y^2+2z^2-6x+2y-4z-25=0

Q5.The direction ratio of the line x-y+z-5=0=x-3y-6 are
  •  3,1,-2
  •  2,-4,1
  •  3/√14,1/√14,(-2)/√14
  •  2/√41,(-4)/√41,1/√41
 If l,m,n are the direction cosines of the line, then 1∙l-1∙m+1∙n=0 and 1∙l-3∙m+0∙n=0 ∴l/(0+3)=m/(1-0)=n/(-3+1) Hence, the direction ratios of the line are 3,1,-2

Q6. The point in the xy-plane which is equidistant from the points (2,0,3), (0,3,2) and (0,0,1) is
  •  (1,2,3)
  •  (3,2,0)
  • (-3,2,0)
  •  (3,-2,0)

Q7.If (2,-1,3) is the foot of the perpendicular drown from the origin to the plane, then the equation of the plane is
  •  2x+y-3z+6=0
  •  2x-y+3z-13=0
  •  2x-y+3z-14=0
  •  2x+y+3z-10=0
Let the equation of any plane passing through P(2,-1,3) is a(x-2)+b(y+1)+c(z-3)=0 …..(i) ∴ DR’s of OP=2,-1,3 Since, the line OP is perpendicular to the plane, therefore the DR’s of the normal to the plane is proportional to the DR’s of OP. ∴ Required equation of plane is 2(x-2)-1(y+1)+3(z-3)=0 ⟹2x-y+3z-14=0

Q8.The points (5, 2, 4), (6, -1, 2) and (8, -7,k) are collinear, if k is equal to
  •  2
  •  -2
  •  3
  •  -1
Let the give points are A,B and C respectively ∴ Direction ratios of AB and BC are 1, -3,-2 and 2,-6,K-2 respectively Since given points are collinear ∴2/1=(-6)/(-3)=(K-2)/(-2) ⇒K-2=-4 ⇒K=-2

Q9.The line passing through the points (5,1,a) and (3,b,1) crosses the yz-plane at the point (0,17/2,-13/2).Then,
  •  a=2,b=8
  •  a=2,b=8
  •  a=4,b=6
  •  a=6,b=4
Equation of the line passing through (5,1,a) and (3,b,1) is (x-3)/(5-3)=(y-b)/(1-b)=(z-1)/(a-1) …..(i) Also,point (0,17/2,-13/2) satisfies Eq.(i),we get -3/2=(17/2-b)/(1-b)=(-13/2-1)/(a-1) From Ist and IIIrd terms a-1=((-15/2))/((-3/2) )⟹a=6 From Ist and IIIed terms-3(1-b)=2(17/2-b)⟹b=4

Q10. A line makes angles of 45° and 60° with the x-axis and the z-axis respectively. The angle made by it with y-axis is
  •  30° or 150°
  •  60° or 120°
  •  45° or 135°
  • 90°
∵ cos^2⁡α+cos^2⁡β+ ⁡cos^2⁡γ=1 ⟹ cos^2 45°+cos^2 β+cos^2 60°=1 ⟹ cos^2 β=1/4 ⟹cos⁡β=±1/2 ⟹ β=60° or 120

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