Three-Dimensional-Geometry-Quiz-4

Three Dimensional Geometry Quiz

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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

Solution

(d) Since line of intersection is perpendicular to both the planes, direction ratios of the line of intersection =|(i ̂j ̂k ̂2 3 1 1 3 2)|=3i ̂-3j ̂+3k ̂ Hence, cos⁡α=3/√(9+9+9)=1/√3

Q4. 

For what value(s) of a, will the two points (1,a,1) and (-3,0,a) lie on opposite sides of the plane 3x+4y-12z+13=0?

•  a<-1 or a>1/3
•  a=0 only
•  0
•  -1

Solution

(a) We must have (3+4a-12+13)(-9-12a+13)<0 ⇒(a+1)(12a-4)>0 ⇒a<-1 or a>1/3

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

Solution
 

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)

Solution

(a) Given lines are (x-5)/3=(y-7)/(-1)=(z+2)/1=r_1 (say) and (x+3)/(-36)=(y-3)/2=(z-6)/4=r_2 (say) ∴x=3r_1+5=-36r_2-3 y=-r_1+7=3+2r_2 and z=r_1-2=4r_2+6 On solving, we get x=21,y=5/3,z=10/3

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

Solution

(a) The plane is perpendicular to the line (x-a)/cos⁡θ =(y+2)/sin⁡θ =(z-3)/0 Hence, the direction ratios of the normal of the plane are cos⁡θ,sin⁡θ, and 0 (i) Now, the required plane passes through the z-axis. Hence the point (0, 0, 0) lies on the plane From Eqs. (i) and (ii), we get equation of the plane as cos⁡θ (x-0)+sin⁡θ (y-0)+0(z-0)=0 cos⁡θ x+sin⁡θ y=0 x+y tan⁡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

Solution

(b) (b) Here, α=β=γ ∵cos^2⁡α+cos^2⁡β+cos^2⁡γ=1 ∴ cos⁡α=1/√3 DC’s of PQ are (1/√3,1/√3,1/√3) PM=Projection of AP on PQ =|(-2+3) 1/√3+(3-5) 1/√3+(1-2) 1/√3|=2/√3 and AP=√((-2+3)^2+(3-5)^2+(1-2)^2 )=√6 AM=√((AP)^2-(PM)^2 )=√(6-4/3)=√(14/3)

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

Solution

(b) Coordinates of L and M are (0,b,c) and (a,0,c), respectively. Therefore, the equation of the plane passing through (0, 0, 0), (0,b,c) and (a,0,c) is |(x-0 y-0 z-0 0 b c a 0 c)|=0 or x/a+y/b-z/c=0