## Three Dimensional Geometry Quiz

Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

Q1. If the lines (x-1)/2=(y+1)/3=(z-1)/4 and (x-3)/1=(y-k)/2=z/1 interested, then the value of k is
•  3/2
•  9/2
•  -2/9
•  -3/2
Solution
(b) Given,(x-1)/2=(y+1)/3=(z-1)/4=λ and (x-3)/1=(y-k)/2=z/1=μ ⟹x=2λ+1,y=3λ-1,z=4λ+1 and x=μ+3,y=2μ+k,z=μ As the lines intersect they must have a point in common. ∴2λ+1=μ+3,3λ-1=2μ+k,4λ+1=μ ⟹ λ=-3/2 and μ=-5 ∴k=3λ-2μ-1 ⟹k=3(-3/2)-2(-5)-1 ⟹k=9/2

Q2.In a three-dimensional xyz space, the equation x^2-5x+6=0 represents
•  Points
•  Planes
•  Curves
•  Pair of straight lines
Solution
(b) x^2-5x+6=0 ⇒x-2=0,x-3=0 Which represents planes

Q3.  The direction ratios of a normal to the plane through (1, 0,0) and (0, 1, 0), which makes an angle of π/4 with the plane x+y=3 are
•   <1,√2,1>
•  <1,1,√2>
•  <1,1,2>
•  <√2,1,1>
Q5.A plane passes through a fixed point (a,b,c). The locus of the foot of the perpendicular to it from the origin is a sphere of radius
•  1/2 √(a^2+b^2+c^2 )
•  √(a^2+b^2+c^2 )
•  a^2+b^2+c^2
•  -1/2(a^2+b^2+c^2)
Q6. Line r =a +λb will not meet the plane r ∙n =q, if
•  b ∙n =0,a ∙n =q
•  b ∙n ≠0,a ∙n ⃗≠q
• b ⃗∙n =0,a ∙n ≠q
•  b ∙n ≠0,a ∙n =q
Q7.Let A(a ) and B(b ) be points on two skew lines r =a +λ p and r =b +uq ⃗ and the shortest distance between the skew lines is 1, where p and q are unit vectors forming adjacent sides of a parallelogram enclosing an area of 1/2 units. If an angle between AB and the line of shortest distance is 60°, then AB=
•  1/2
•  2
•  1
•  λ∈R-{0}
Q8.Consider triangle AOB in the x-y plane, where A≡(1,0,0);B≡(0,2,0); and O≡(0,0,0). The new position of O, when triangle is rotated about side AB by 90° can be
•  (4/5,3/5,2/√5)
•  T((-3)/5,√2/5,2/√5)
•  (4/5,2/5,5/√5)
•  (4/5,2/5,1/√5)
Solution
(c) Equation of a line AB is (x-1)/1=y/(-2)=z/0=λ Now AB⊥OC⇒ 1(λ+1)+(-2λ)(-2)=0 ⇒5λ=-1 ⇒ λ=-1/5 ⇒C is (4/5,2/5,0). Now x_1^2+(y_1-2)^2+z_1^2=4 (i) and (x_1-1)^2+y_1^2+z_1^2=1 (ii) Now OC⊥CD ⇒ (x_1-4/5) 4/5+(y_1-2/5) 2/5+(z_1-0)0=0 (iii) From (i) and (ii), we get -4y_1+2x_1=0 ⇒ x_1=2y_1 From (iii), putting x_1=2y_1⇒2y_1=4/5 ⇒y_1=2/5 ⇒x_1=4/5. Putting this value of x_1 and y_1 in (i), we get z_1=±2/√5

Q9.Let A(1,1,1),B(2,3,5) and C(-1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is
•  2x-3y+z+2√14=0
•  2x-3y+z-√14=0
•  2x-3y+z+2=0
•  2x-3y+z-2=0
Q10. Value of λ such that the line (x-1)/2=(y-1)/3=(z-1)/λ is ⊥ to normal to the plane r ∙(2i + 3j + 4k )=0 is
•  -13/4
•  -17/4
•  4
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