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Q1. The shortest distance between the lines (x-2)/3=(y+3)/4=(z-1)/5 and (x-5)/1=(y-1)/2=(z-6)/3 is
  •  3
  •  2
  •  1
  •  0
Solution
Shortest distance =



Q2. The equation of the plane through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x+6y+6z-1=0, is
  •  3x+4y+5z=9
  •  3x+4y-5z=9
  •  3x+4y-5z-9=0
  •   None of these
Solution
B

Q3.  The equation of the plane containing the line (x+1)/-3=(y-3)/2=(z+2)/1 and the point 0, 7, -7, is
  •   x+y+z=1
  •  x+y+z=2
  •  None of these
  •  x+y+z=0
Solution
The position vector of any point on the given line is i ̂+j ̂+λ(2i ̂+j ̂+4k ̂ )=(2λ+1) i ̂+(λ+1) j ̂+4 λ k ̂ Clearly, this point lies on the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 Hence, the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 contains the given line

Q4. f the direction ratio of two lines are given by 3lm-4ln+mn=0 and l+2m+3n=0, then the angle between the line is
  •  π/6
  •  π/4
  •  π/3
  •  π/2
Solution
Given, 3lm-4ln+mn=0 …..(i) and l+2m+3n=0 …(ii) From Eq. (ii), l=-(2m+3n) putting in Eq. (i) -3(2m+3n)m+4(2m+3n)n+mn=0 ⟹ -〖6m〗^2+〖12n〗^2=0 ⟹m=±√2 n Now,m=√2 n ⟹l=-(2√2 n+3n)=-(2√2+3)n ∴l:m:n=-(3+2√2)n:√2 n:n =-(3+2√2):√2:1 Also,m=-√2 n⟹l=-(-2√2+3)n ∴l:m:n=-(3-2√2)n:-√2:n =-(3-2√2):-√2:1 =cos⁡θ =((3+2√2)(3-2√2)+(√2)(-√2)+1∙1)/(√((3+2√2 )^2+(√2)^2+1^2 ) √((3-2√2)^2+(-√2)^2+1^2 )) =0 ⟹ θ=π/2

Q5.The planex2+y3+z4=1 cuts the coordinate axes in A,B,C then the area of the ∆ABC
  •  √61 sq units
  •  √41 sq units
  •  √29 sq units
  •   None of these
Solution
 Equation of the plane is x/2+y/3+z/4=1 Here,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 √(2^2∙3^2+3^2∙4^2+4^2∙2^2 ) =1/2 √244=√61 sq units

Q6. The equation to the straight line passing through the points (4, -5, -2) and (-1, 5, 3) is
  •  (x+1)/1=(y-5)/2=(z-3)/(-1)
  •  (x-4)/1=(y+5)/(-2)=(z+2)/(-1)
  • x/(-1)=y/5=z/3
  •  x/4=y/(-5)=z/(-2)
Solution

 


Q7.A straight line r=a+λ b meets the plane rn=0 in P. The position vector of P is
  •  a ⃗+[(a ⃗∙n ⃗)/(b ⃗∙n ⃗ )] b ⃗

  •  a ⃗-[(a ⃗∙n ⃗)/(b ⃗∙n ⃗ )] b ⃗
  •  a ⃗-[(a ⃗∙n ⃗)/(b ⃗∙n ⃗ )] b ⃗
  •  None of these
  • Solution
    The straight line r ⃗=a ⃗+λ b ⃗ meets the plane r ⃗∙n ⃗=0 in P for which λ is given by (a ⃗+λ b ⃗ )∙n ⃗=0⇒λ=-(a ⃗∙n ⃗)/(b ⃗∙n ⃗ ) Thus, the position vector of P is r ⃗=a ⃗((a ⃗∙n ⃗)/(b ⃗∙n ⃗ )) b ⃗ [Putting the value of λ in r ⃗=a ⃗+λb ⃗ ]

    Q8.Cosine of the angle between two diagonals of cube is equal to
    •  2/√6
    •  1/3
    •  1/2
    •  None of these
    Solution
    Let OA,OB,OC be the sides of a cube such that OA=OB=OC=a ∴ Direction ratios of OE are (a-0,a-0)ie (a,a,a) ∴Direction cosines of AF are (1/√3,1/√3,1/√3) Similarly, direction of AF are (-1/√3,1/√3,1/√3). ∴ Angle between OE and AF is cos^(-1)⁡〖[-1/√3∙1/√3+1/√3∙1/√3+1/√3∙1/√3] cos^(-1)⁡(1/3) 〗

    Q9.The line segment adjoining the points A, B makes projection 1, 4, 3 on x, y, z-axes respectively. Then, the direction cosines of AB are
    •  1, 4, 3
    •  1/√26,-4/√26,3/√26
    •  -1/√26,4/√26,3/√26
    •  1/√26,4/√26,3/√26
    Solution
    DC^' s of AB=1/√(1^2+4^2+3^2 ),4/√(1^2+4^2+3^2 ),3/√(1^2+4^2+3^2 ) =1/√26,4/√26,3/√26

    Q10.  A variable plane moves so that sum of the reciprocals of its intercepts on the coordinate axes is 1/2 Then, the plane passes through
    •  (1/2,1/2,-1/2)
    •  (2, 2, 2)
    •  (-1,1,1)
    • (0, 0, 0)
    Solution
    Let x/a+y/b+z/c=1 ….(i) Then,1/a+1/b+1/c=1/2 …(ii) On comparing Eqs. (i) and (ii), we get x=2,y=2,z=2

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