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3D Geometry question

2021 · 24 Feb · Shift 2 · Q25
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  5. /2021 · 24 Feb · Shift 2 · Q25

3D Geometry question

2021 · 24 Feb · Shift 2 · Q25

JEE MainMathematics3D GeometryMCQ+4 / −1
Let a, b ∈\in∈ R. If the mirror image of the point P(a, 6, 9) with respect to the line x−37=y−25=z−1−9{{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}7x−3​=5y−2​=−9z−1​ is (20, b, −-− a −-− 9), then | a + b |, is equal to :
  1. A
    88
  2. B
    90
  3. C
    86
  4. D
    84
View written solutionFree

Correct answer: A

  1. Interpret the line of reflection

The given line is

x−37=y−25=z−1−9.\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}.7x−3​=5y−2​=−9z−1​.

So it passes through A=(3,2,1)A=(3,2,1)A=(3,2,1) and has direction vector d⃗=(7,5,−9).\vec d=(7,5,-9).d=(7,5,−9).

  1. Use the fact about mirror image with respect to a line in 3D

If a point PPP is reflected in a line, then:

  • the midpoint of PPP and its image P′P'P′ lies on the line,
  • the segment PP′PP'PP′ is perpendicular to the line.

Given P=(a,6,9),P′=(20,b,−a−9).P=(a,6,9), \qquad P'=(20,b,-a-9).P=(a,6,9),P′=(20,b,−a−9).

Let the midpoint be MMM. Then

M=(a+202,6+b2,9+(−a−9)2)=(a+202,6+b2,−a2).M=\left(\frac{a+20}{2},\frac{6+b}{2},\frac{9+(-a-9)}{2}\right) =\left(\frac{a+20}{2},\frac{6+b}{2},-\frac a2\right).M=(2a+20​,26+b​,29+(−a−9)​)=(2a+20​,26+b​,−2a​).
  1. Condition 1: PP′PP'PP′ is perpendicular to the line

Vector

PP′→=(20−a, b−6, −a−18).\overrightarrow{PP'}=(20-a,\,b-6,\,-a-18).PP′=(20−a,b−6,−a−18).

Since this is perpendicular to the line direction (7,5,−9)(7,5,-9)(7,5,−9),

(20−a, b−6, −a−18)⋅(7,5,−9)=0.(20-a,\,b-6,\,-a-18)\cdot(7,5,-9)=0.(20−a,b−6,−a−18)⋅(7,5,−9)=0.

Compute:

7(20−a)+5(b−6)−9(−a−18)=07(20-a)+5(b-6)-9(-a-18)=07(20−a)+5(b−6)−9(−a−18)=0 140−7a+5b−30+9a+162=0140-7a+5b-30+9a+162=0140−7a+5b−30+9a+162=0 272+2a+5b=0.272+2a+5b=0.272+2a+5b=0.

So,

2a+5b=−272....(1)2a+5b=-272. \qquad ...(1)2a+5b=−272....(1)
  1. Condition 2: midpoint lies on the line

A general point on the line is

(3+7t, 2+5t, 1−9t).(3+7t,\,2+5t,\,1-9t).(3+7t,2+5t,1−9t).

So for midpoint MMM,

a+202=3+7t,6+b2=2+5t,−a2=1−9t.\frac{a+20}{2}=3+7t, \qquad \frac{6+b}{2}=2+5t, \qquad -\frac a2=1-9t.2a+20​=3+7t,26+b​=2+5t,−2a​=1−9t.

From the third equation,

−a2=1−9t⇒a=−2+18t.-\frac a2=1-9t \Rightarrow a=-2+18t.−2a​=1−9t⇒a=−2+18t.

From the first equation,

a+202=3+7t⇒a+20=6+14t⇒a=−14+14t.\frac{a+20}{2}=3+7t \Rightarrow a+20=6+14t \Rightarrow a=-14+14t.2a+20​=3+7t⇒a+20=6+14t⇒a=−14+14t.

Equating the two expressions for aaa,

−2+18t=−14+14t-2+18t=-14+14t−2+18t=−14+14t 12=−4t12=-4t12=−4t t=−3.t=-3.t=−3.

Then

a=−2+18(−3)=−2−54=−56.a=-2+18(-3)=-2-54=-56.a=−2+18(−3)=−2−54=−56.

Now use the second midpoint equation:

6+b2=2+5(−3)=2−15=−13.\frac{6+b}{2}=2+5(-3)=2-15=-13.26+b​=2+5(−3)=2−15=−13.

Thus,

6+b=−26⇒b=−32.6+b=-26 \Rightarrow b=-32.6+b=−26⇒b=−32.
  1. Compute the required value
a+b=−56+(−32)=−88.a+b=-56+(-32)=-88.a+b=−56+(−32)=−88.

Hence,

∣a+b∣=88.|a+b|=88.∣a+b∣=88.
  1. Check with stored answer

Derived answer is 88, which matches option A.

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