## 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. For the line (x-1)/1=(y-2)/2=(z-3)/3, which one of the following is incorrect?
•  It lies in the plane x-2y+z=0
•  It is same as line x/1=y/2=z/3
•  It passes through (2, 3, 5)
•  It is parallel to the plane x-2y+z-6=0
Solution
(c) (1, 2, 3) satisfies the plane x-2y+z=0 and also (i ̂+2j ̂+3k ̂)∙(i ̂-2j ̂+k ̂ )=0 Since the lines (x-1)/1=(y-2)/2=(z-3)/3 and x/1=y/2=z/3 both satisfy (0, 0, 0) and (1, 2, 3), both are same. Given line is obviously parallel to the plane x-2y+z=6

Q2.The equation of a plane which passes through the point of intersection of lines (x-1)/3=(y-2)/1=(z–3)/2, and (x-3)/1=(y-1)/2=(z-2)/3 and at greatest distance from point (0, 0, 0) is
•  4x+3y+5z=25
•  4x+3y=5z=50
•  3x+4y+5z=49
•  x+7y-5z=2
Solution
(b) Let a point (3λ+1,λ+2,2λ+3) of the first line also lies on the second line Then (3λ+1-3)/1=(λ+2-1)/2=(2λ+3-2)/3⇒λ=1 Hence, the point of intersection P of the two lines (4, 3, 5) Equation of plane perpendicular to OP, where O is (0, 0, 0) and passing through P is 4x+3y+5z=50

Q3.  Let L_1 be the line r _1=2i ̂+j ̂-k ̂+λ(i ̂+2k ̂ ) and let L_2 be the line r _2=3i ̂+j ̂+μ(i ̂+j ̂-k ̂). Let π be the plane which contains the line L_1 and is parallel to L_2. The distance of the plane π from the origin is
•    √(2/7)
•  1/7
•  √6
•  None
•  a ×c =b ×c
•  a ∙c =b ∙c
•  b ×a =c ×a
•  None of these
Q5.The projection of the line (x+1)/(-1)=y/2=(z-1)/3 on the plane x-2y+z=6 is the line of intersection of this plane with the plane
•   2x+y+2=0
•  3x+y-z=2
•   2x-3y+8z=3
•  None of
Solution
(a) Equation of the plane through (-1,0,1) is a(x+1)+b(y-0)+c(z-1)=0 (i) Which is parallel to the given line and perpendicular to the given plane -a+2b+3c=0 (ii) and a-2b+c=0 (iii) From Eqs. (ii) and (iii), we get c=0,a=2b From Eq., 2b(x+1)+by=0 ⇒2x+y+2=0

Q6. A plane makes intercepts OA,OB and OC whose measurements are b and c on the OX,OY and OZ axes. The area of triangle ABC is
•  a) 1/2(ab+bc+ca)
•  b) 1/2 abc(a+b+c)
• 1/2 (a^2 b^2+b^2 c^2+c^2 a^2 )^(1/2)
•  1/2 (a+b+c)^2
Q7.Given α =3i ̂+j ̂+2k ̂ and β =i ̂-2j ̂-4k ̂ are the position vectors of the points A and B. Then the distance of the point -i ̂+j ̂+k ̂ from the plane passing through B and perpendicular to AB is
•  5
•  10
•  15
•  20
Q8.If the distance of the point P(1,-2,1) from the plane x+2y-2z=α, where α>0, is 5, then the foot of the perpendicular from P to the plane is
•  (8/3,4/3,-7/3)
•  T(4/3,-4/3,1/3)
•  (1/3,2/3,10/3)
•  (2/3,-1/3,5/2)
Solution
Distance of point P from plane=5 ∴5|(1-4-2-α)/3| α =10 Foot perpendicular (x-1)/1=(y+2)/2=(z-1)/(-2)-((1-4-2-10))/(1+4+4)=5/3 ⟹x=8/3,y=4/3,z-7/3 Thus, the foot of the perpendicular is A(8/3,4/3,-7/3)

Q9.Shortest distance between the lines (x-1)/1=(y-1)/1=(z-1)/1 and (x-2)/1=(y-3)/1=(z-4)/1 is equal to
•  √14
•  √7
•  √2
•  None of these
Q10. A line with positive direction cosines passes through the point P(2,-1,2) and makes equal angles with the coordinate axes. The line meets the plane 2x+y+z=9 at point Q. The length of the line segment PQ equals
•  1
•  √2
•  √3
• 2
Solution
(c) Since,l=m=n=1/√3 ∴Equation of line is (x-2)/(1/√3)=(y+1)/(1/√3)=(z-2)/(1/√3) ⟹x-2=y+1=z-2=r [say] ∴ Any point on the line is Q=(r+2,r-1,r+2) ∵Q lies on the plane 2x+y+z=9 ∴2(r+2)+(r-1)+(r+2)=9 ⟹4r+5=9 ⟹r=1 ∴ Coordinate Q(3,0,3) ∴ PQ=√(〖(3-2)〗^2+〖(0+1)〗^2+〖(3-2)〗^2 )=√3 #### Written by: AUTHORNAME

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