## 3-D Coordinate Geometry Quiz-11

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced..

Q1. A plane makes intercepts a,b,c at A,B,C on the coordinate axes respectively.
If the centroid of the ∆ABC is at (3,2,1)then the equation of the plane is :
•  x+2y+3z=9
•  2x-3y-6z=18
•  2x+3y+6z=18
•  2x+y+6z=18
Solution
(c) Equation of plane is

x/a+y/b+z/c=1

Also,3=(a+0+0)/3

⟹a=9 and similarly b=6 and c=3

∴ Equation of required plane is

x/9+y/6+z/3=1

⟹2x+3y+6z=18

Q2.A line with direction ratios proportional to 2, 1, 2 meets each of the lines x=y+a=z and x+a=2y=2z. The coordinates of the points of intersection are given by :
•  (3a,3a,3a),(a,a,a)
•  (3a,2a,3a),(a,a,a)
•  (3a,2a,3a),(a,a,2a)
•  (2a,3a,3a),(2a,a,a)
Solution
(b) Let the equation of line AB be

(x-0)/1=(y+a)/1=(z-0)/1=k [say]

∴ Coordinate of E are (k,k-a,k).

Also, the equation of the other line CD is

(x+a)/2=(y-0)/1=(z-0)/1=λ [say]

∴ Coordinates of F are (2λ-a,λ,λ)

Direction ratio of FE are {(k-2λ+a),(k-λ-a),(k-λ)}

∴ (k-2λ+a)/2=(k-λ-a)/1=(k-λ)/2

From Ist and IInd terms,

k-2λ+a=2k-2λ-2a

⟹k=3a

From IInd and IIIrd terms,

2k-2λ-2a=k-λ

⟹ λ=k-2a=3a-2a

⟹ λ=a

∴ Coordinate of E=(3a,2a,3a) and coordinate of F=(a,a,a)

Q3.  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

Q4. A line joining points (4,-1,2) and (-3,2,3) meets the plane at the point (-10,5,4) at 90°, then equation of the plane is:
•  7x-3y-z+89=0
•  7x+3y+z+89=0
•  7x-3y+z+89=0
•  None of these
Solution
(a) ∵ A line joining points (4,-1,2) and (-3,2,3) meets the plane at 90°, then this line is normal to the plane

Also, DR’s of normal are <-7,3,1>

∴ DR’s of plane are <-7,3,1>and point (-10,5,4) lies on the plane

Hence, equation of plane is

-7(x+10)+3(y-5)+1(z-4)=0

⇒7x-3y-z+89=0

Q5. If for a plane, the intercepts on the coordinate axes are 8, 4, 4, then the
length of the perpendicular from the origin to the plane is:
•  8/3
•  3/8
•  3
•  d) 4/3
Solution
(a) Equation of plane is x/8+y/4+z/4=1

⇒x+2y+2z=8

Length of perpendicular from origin to the plane

x+2y+2z-8=0 is

|(-8)/√(1+4+4)|=8/3

Q6. A straight line which makes an angle of 60° with each of y and z-axes, this line makes with x-axis at an angle:
•  30°
•  60°
• 75°
•  45°
Solution
(d) Clearly, cos2 α+cos2 60°+cos2 60°=1 where α is the angle which the straight line makes with x-axis

∴ cos2 α=1-1/4-1/4=1/2

⟹cos⁡〖α=1/√2⟹α=45°〗

Q7. The equation of the sphere touching the three coordinate planes is:
•  x2+y2+z2+2a(x+y+z)+2a2=0
•  x2+y2+z2-2a(x+y+z)+2a2=0
•  x2+y2+z2±2a(x+y+z)+2a2=0
•  x2+y2+z2±2ax±2ay±2az+2a2=0
Solution
(b)
Since, the given sphere touching the three coordinates planes. So, it is clear that centre is (a,a,a) and radius is a

∴ The equation of sphere at the centre (a,a,a) and radius a is

(x-a)2+(y-a)2+(z-a)2=a2

⇒x2+y2+z2-2ax-2ay-2az+3a2=a2

∴x2+y2+z2-2a(x+y+z)+2a2=0 is the required equation of sphere

Q8. The equation of the plane passing through the mid-point of the line segment of join of the points P(1,2,3)
and Q(3,4,5) and perpendicular to it is :
•  x+y+z=9
•  x+y+z=-9
•  2x+3y+4z=9
•  2x+3y+4z=-9
Solution
(a)
The coordinates of the mid-point of PQ are (2, 3, 4). The direction ratios of PQ are proportional to 3-1,4-2,5-3 i.e. 1, 1, 1

So, equation of the required plane is

1×(x-2)+1×(y-3)+1×(z-4)=0 or, x+y+z=9

Q9. Angle between the line (x+1)/1=y/2=(y-1)/1 and a normal to the plane x-y+z=0 is:
•
•  30°
•  45°
•  90°
Solution
(a)
∵ sin⁡〖θ=(a_1 a_2+b_1 b_2+c_1 c_2)/(√(a_12+b_12+c_12 ) √(a_22+b_22+c_22 ))〗

=(1×1+2×-1+1×1)/(√(1+4+1) √(1+1+1))=0

⟹ θ=0°

Q10. If from a point P(a,b,c) perpendiculars PA,PB are drown to yz and zx plane, then the equation of the plane OAB is
•  bcx+cay+abz=0
•  bcx+cay-abz=0
•  bcx-cay+abz=0
• -bcx+cay+abz=0
Solution
(b)
Since, PA,PB are perpendicular drawn from P(a,b,c) on yz and zx-planes.

∴ A(0,b,c) and B(a,0,c) are the points on yz and xz-plances.

The equation of plane passing through (0, 0, 0) is

Ax+By+Cz=0

Which also passes through points A and B

∴A∙0+B∙b+C∙c=0

and A∙a+B∙0+C∙c=0

⟹A/(bc-0)=B/(ac-0)=C/(0-ab)=λ [say]

⟹A=λbc,B=λac,C=-λab

∴ Required equation is bcx+acy-abz=0 ## Want to know more

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