## 3D Coordinate Geometry Quiz-14

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..
3D Coordinate Geometry Quiz-14
Q1. If the angle between the line  $$\frac{(x+1)}{1}=\frac{(y-1)}{2}=\frac{(z-2)}{2}$$ and the plane 2x-y+√λ z+4=0 is such that sin⁡θ=1/3 Then, value of λ is:
• -4/3
•  3/4
• -3/5
•  5/3
Solution

Q2.Let the line$$\frac{(x-2)}{3}=\frac{(y-1)}{-5}=\frac{(z+2)}{2}$$ lies in the plane x+3y-αz+β=0 Then (α,β) equals:
•  (6,-17)
•  (-6,7)
•  (5,-15)
•  (-5,15)
Solution
(b)
DR’s of given line are (3,-5,2)
DR’s of normal to the plane =(1,3,-α)
∴ Line is perpendicular to the normal
⟹3(1)-5(3)+2(-α)=0
⟹3-15-2α=0
⟹2α=-12
⟹ α=-6
Also point (2,1,-2) lies on the plane

2+3+6(-2)+β=0
⟹ β=7
∴(α,β)=(-6,7)

Q3.  The distance between the line $\stackrel{\to }{\mathrm{r.}}$=2$\stackrel{^}{i}$-2$\stackrel{^}{j}$ +3$\stackrel{^}{k}$ +λ($\stackrel{^}{i}$ -$\stackrel{^}{j}$ +4$\stackrel{^}{k}$) and the plane $\stackrel{\to }{\mathrm{r.}}$($\stackrel{^}{i}$+5$\stackrel{^}{j}$ +$\stackrel{^}{k}$)=5, is:
•  10/(3√3)
•  10/3
•  10/9
•  None of these
Solution

Q4. In a three dimensional xyz-space, the equation x2-5x+6=0 represents:
•  Points
•  Plane
•  Curves
•  Pair of straight lines
Solution
(b)
Given equation is x2 -5x+6=0
⇒(x-2)(x-3)=0
⇒(x-2)=0 or (x-3)=0
Which represents a plane

Q5. The direction ratios of a normal to the plane passing through (0, 0, 1), (0, 1, 2) and (1, 2, 3) are proportional to:
•  0,1,-1
•  1,0,-1
•  0,0,-1
•  1, 0, 0
Solution

Q6. The image of the point (1, 2, 3) in lie  $$\frac{x}{2}=\frac{(y-1)}{3}=\frac{(z-1)}{3}$$is
•  (1,5/2,5/2)
•  (1,9/4,11/4)
• (1, 3, 2)
•  (3, 1, 2)
Solution

Q7. The equation of the plane passing through (1, 1, 1) and (1,-1,-1) and perpendicular to 2x-y+z+5=0 is :
•  2x+5y+z-8=0
•  x+y-z-1=0
•  2x+5y+z+4=0
•  x-y+z-1=0
Solution

Q8.If a sphere of radius r passes through the origin, then the extremities of the diameter parallel to x-axis lie on each of the spheres:
•  x2+y2+z2±2rx=0
•  x2+y2+z2±2ry=0
•  x2+y2+z2±2rz=0
•  x2+y2+z2±2ry±2rz=0
Solution
(a)
Let (u,v,w) be the centre of the sphere with radius r. Since, it passes through the origin
∴u2 +v2 +w2 =r2 . Equation of the diameter parallel to x-axis is
$$\frac{(x-u)}{1}=\frac{(y-v)}{0}=\frac{(z-w)}{0}$$ …(i)
As it passes through u,v,w and direction ratios of x- axis are 1, 0, 0
The extremities of diameter are the points on Eq. (i) at a distance r from the centre (u,v,w)

∴ The required extremities are P(r+u,v,w) and Q(-r+u,v,w)
P lies on the sphere x2 +y2 +z2 -2rx=0 as
(r+u)2 +v2 +w2 -2r(r+u)=0
Because u2 +v2 +w2 =r2
and similarly Q lies on the sphere x2 +y2 +z2 +2rx=0

Q9. A And B are two give points. Let C divides AB internally and D divides AB externally in the same ratio. Then AC,AB,AD are in:
•  AP
•  GP
•  HP
•  None of these
Solution
Not Available

Q10. The length of the perpendicular from the origin to the plane passing through the point $\stackrel{\to }{a}$ and containing the line $\stackrel{\to }{r}$ =$\stackrel{\to }{b}$$\stackrel{\to }{c}$ is:
•  [$\stackrel{\to }{a}$ $\stackrel{\to }{b}$ $\stackrel{\to }{c}$ ]/|$\stackrel{\to }{a}$ ×$\stackrel{\to }{b}$ +$\stackrel{\to }{b}$ ×$\stackrel{\to }{c}$ +$\stackrel{\to }{c}$ ×$\stackrel{\to }{a}$ |
•  [$\stackrel{\to }{a}$ $\stackrel{\to }{b}$ $\stackrel{\to }{c}$ ]/|$\stackrel{\to }{a}$ ×$\stackrel{\to }{b}$ +$\stackrel{\to }{b}$ ×$\stackrel{\to }{c}$ |
•  [$\stackrel{\to }{a}$ $\stackrel{\to }{b}$ $\stackrel{\to }{c}$ ]/|$\stackrel{\to }{b}$ ×$\stackrel{\to }{c}$ +$\stackrel{\to }{c}$ ×$\stackrel{\to }{a}$ |
• [$\stackrel{\to }{a}$ $\stackrel{\to }{b}$ $\stackrel{\to }{c}$ ]/|$\stackrel{\to }{c}$ ×$\stackrel{\to }{a}$ +$\stackrel{\to }{a}$ ×$\stackrel{\to }{b}$ |
Solution

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