## 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. The length of the perpendicular drawn from (1, 2, 3) to the line (x-6)/3=(y-7)/2=(z–7)/(-2) is
•  4
•  5
•  6
•  7
Solution
(d) Let P be the point (1,2,3) and PN be the length of the perpendicular from P on the given line Coordinates of point N are (3λ+6,2λ+7,-2λ+7) Now PN is perpendicular to the given line or vector 3i ̂+2j ̂-2k ̂ ⇒3(3λ+6-1)+2(2λ+7-2)-2(-2λ+7-3)=0 ⇒ λ=-1 Then, point N is (3, 5, 9) ⇒PN=7

Q2.If angle θ between the line (x+1)/1=(y-1)/2=(z-2)/2 and the plane 2x-y+√λ z+4=0 is such that sin⁡θ=1/3, the value of λ is
•  -3/5
•  5/3
•  -4/3
•  3/4
Solution
(b) The line is (x+1)/1=(y-1)/2=(z-2)/2 and the plane is 2x-y+√λ z+4=0 If θ be the angle between the line and the plane, then 90°-θ is the angle between the line and normal to the plane ⇒cos⁡(90°-θ)=((1)(2)+(2)(-1)+(2)(√λ))/(√(1+4+4) √(4+1+λ)) ⇒sin⁡θ=(2-2+2√λ)/(3√(5+λ)) ⇒1/3=(2√λ)/(3√(5+λ)) ⇒√(5+λ)=2√λ ⇒5+λ=4λ ⇒3λ=5 ⇒ λ=5/3

Q3.  Let L be the line of intersection of the planes 2x+3y+z=1 and x+3y+2z=2. If L makes an angle α with the positive x-axis, then cos⁡α equals
•   1/2
•  1
•  1/√2
•  1/√3
•  a<-1 or a>1/3
•  a=0 only
•  0
•  -1
Q5.The reflection of the point a in the plane r ∙n =q is
•   a +((q -a ∙n ))/(|n |)
•  a+2(((q-a ∙n))/|n|^2 ) n ⃗
•   a+(2(q+a∙n))/(|n|)n
•  None of these
Q6. The point of intersection of the lines (x-5)/3=(y-7)/(-1)=(z+2)/1 and (x+3)/(-36)=(y-3)/2=(z-6)/4 is
•  (21,5/3,10/3)
•  (2,10,4)
• (-3,3,6)
•  (5,7,-2)
Q7.What is the equation of the plane which passes through the z-axis and is perpendicular to the line (x-a)/cos⁡θ =(y+2)/sin⁡θ =(z-3)/0?
•  x+y tan⁡θ=0
•  y+x tan⁡θ=0
•  x cos⁡θ-y sin⁡θ=0
•  x sin⁡θ-y cos⁡θ=0
Q8.The line (x+6)/5=(y+10)/3=(z+14)/8 is the hypotenuse of an isosceles right angled triangle whose opposite vertex is (7,2,4). Then which of the following is not the side of the triangle?
•  (x-7)/2=(y-2)/(-3)=(z-4)/6
•  T(x-7)/3=(y-2)/6=(z-4)/2
•  (x-7)/3=(y-2)/5=(z-4)/(-1)
•  None of these
Solution
(c) Given one vertex A(7,2,4) and line (x+6)/5=(y+10)/3=(z+14)/8 General point on above line B≡(5λ-6,3λ-10,8λ-14) Direction ratios of line AB are <5λ-13,3λ-12,8λ-18> Direction ratios of line BC are <5,3,8> Since angle between AB nad BC is π/4 cos⁡〖π/4〗=((5λ-3)5+3(3λ-12)+8(8λ-18))/(√(5^2+3^2+8^2 )∙√((5λ-13)^2+(3λ-12)^2+(8λ-18)^2 )) Squaring and solving, we have λ=3,2 Hence equation of lines are (x-7)/2=(y-2)/(-3)=(z-4)/6 and (x-7)/3=(y-2)/6=(z-4)/2

Q9.The distance of point A(-2,3,1) from the line PQ through P(-3,5,2), which makes equal angles with the axes is
•  2/√3
•  √(14/3)
•  16/√3
•  5/√3
Q10. From the point P(a,b,c), let perpendicular PL and PM be drawn to YOZ and ZOX planes, respectively. Then the equation of the plane OLM is
•  x/a+y/b+z/c=0
•  x/a+y/b-z/c=0
•  x/a-y/b-z/c=0
• x/a-y/b+z/c=0 #### Written by: AUTHORNAME

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