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3D Geometry question

2022 · 27 Jul · Shift 2 · Q32
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  5. /2022 · 27 Jul · Shift 2 · Q32

3D Geometry question

2022 · 27 Jul · Shift 2 · Q32

JEE MainMathematics3D GeometryMCQ+4 / −1
If the length of the perpendicular drawn from the point P(a,4,2)P(a, 4,2)P(a,4,2), a >0\gt 0>0 on the line x+12=y−33=z−1−1\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}2x+1​=3y−3​=−1z−1​ is 262 \sqrt{6}26​ units and Q(α1,α2,α3)Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)Q(α1​,α2​,α3​) is the image of the point P in this line, then a+∑i=13αi\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}a+i=1∑3​αi​ is equal to :
  1. A
    7
  2. B
    8
  3. C
    12
  4. D
    14
View written solutionFree

Correct answer: B

  1. Write the line in vector/parametric form

The line is

x+12=y−33=z−1−1=t\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=t2x+1​=3y−3​=−1z−1​=t

So a general point on the line is

(x,y,z)=(−1+2t, 3+3t, 1−t)(x,y,z)=(-1+2t,\ 3+3t,\ 1-t)(x,y,z)=(−1+2t, 3+3t, 1−t)

A point on the line is A(−1,3,1)A(-1,3,1)A(−1,3,1) and its direction vector is d⃗=(2,3,−1).\vec d=(2,3,-1).d=(2,3,−1).

The given point is P(a,4,2).P(a,4,2).P(a,4,2).

  1. Use the distance from a point to a line

Distance from point PPP to the line through AAA with direction d⃗\vec dd is

distance2=∣AP⃗∣2−(AP⃗⋅d⃗)2∣d⃗∣2.\text{distance}^2=|\vec{AP}|^2-\frac{(\vec{AP}\cdot \vec d)^2}{|\vec d|^2}.distance2=∣AP∣2−∣d∣2(AP⋅d)2​.

Here,

AP⃗=P−A=(a+1,1,1).\vec{AP}=P-A=(a+1,1,1).AP=P−A=(a+1,1,1).

Also,

∣d⃗∣2=22+32+(−1)2=14.|\vec d|^2=2^2+3^2+(-1)^2=14.∣d∣2=22+32+(−1)2=14.

Now,

∣AP⃗∣2=(a+1)2+12+12=(a+1)2+2.|\vec{AP}|^2=(a+1)^2+1^2+1^2=(a+1)^2+2.∣AP∣2=(a+1)2+12+12=(a+1)2+2.

And

AP⃗⋅d⃗=2(a+1)+3(1)−1(1)=2a+4.\vec{AP}\cdot \vec d=2(a+1)+3(1)-1(1)=2a+4.AP⋅d=2(a+1)+3(1)−1(1)=2a+4.

Given perpendicular distance is 262\sqrt 626​, so

(26)2=24.(2\sqrt 6)^2=24.(26​)2=24.

Hence,

(a+1)2+2−(2a+4)214=24.(a+1)^2+2-\frac{(2a+4)^2}{14}=24.(a+1)2+2−14(2a+4)2​=24.

Multiply by 141414:

14(a+1)2+28−(2a+4)2=336.14(a+1)^2+28-(2a+4)^2=336.14(a+1)2+28−(2a+4)2=336.

Expand:

14(a2+2a+1)+28−(4a2+16a+16)=33614(a^2+2a+1)+28-(4a^2+16a+16)=33614(a2+2a+1)+28−(4a2+16a+16)=336 14a2+28a+14+28−4a2−16a−16=33614a^2+28a+14+28-4a^2-16a-16=33614a2+28a+14+28−4a2−16a−16=336 10a2+12a+26=33610a^2+12a+26=33610a2+12a+26=336 10a2+12a−310=010a^2+12a-310=010a2+12a−310=0 5a2+6a−155=0.5a^2+6a-155=0.5a2+6a−155=0.

Solve:

5a2+31a−25a−155=05a^2+31a-25a-155=05a2+31a−25a−155=0 a(5a+31)−5(5a+31)=0a(5a+31)-5(5a+31)=0a(5a+31)−5(5a+31)=0 (a−5)(5a+31)=0.(a-5)(5a+31)=0.(a−5)(5a+31)=0.

So

a=5ora=−315.a=5 \quad \text{or} \quad a=-\frac{31}{5}.a=5ora=−531​.

Given a>0a>0a>0, therefore

a=5.a=5.a=5.
  1. Find the foot of the perpendicular / image point on the line

Since Q(α1,α2,α3)Q(\alpha_1,\alpha_2,\alpha_3)Q(α1​,α2​,α3​) is the image of PPP in the line, this means the reflection of PPP about the line.
So first find the foot of perpendicular MMM of PPP on the line; then MMM is the midpoint of PQPQPQ.

Let

M=(−1+2t, 3+3t, 1−t)M=(-1+2t,\ 3+3t,\ 1-t)M=(−1+2t, 3+3t, 1−t)

be the foot on the line.

Condition for perpendicularity:

PM→⋅d⃗=0.\overrightarrow{PM}\cdot \vec d=0.PM⋅d=0.

Now with a=5a=5a=5, point P=(5,4,2)P=(5,4,2)P=(5,4,2).

So

AM→=td⃗,\overrightarrow{AM} = t\vec d,AM=td,

and parameter of projection is

t=AP⃗⋅d⃗∣d⃗∣2.t=\frac{\vec{AP}\cdot \vec d}{|\vec d|^2}.t=∣d∣2AP⋅d​.

We already have for a=5a=5a=5,

AP⃗=(6,1,1),\vec{AP}=(6,1,1),AP=(6,1,1), AP⃗⋅d⃗=6⋅2+1⋅3+1⋅(−1)=12+3−1=14.\vec{AP}\cdot \vec d=6\cdot2+1\cdot3+1\cdot(-1)=12+3-1=14.AP⋅d=6⋅2+1⋅3+1⋅(−1)=12+3−1=14.

Thus,

t=1414=1.t=\frac{14}{14}=1.t=1414​=1.

Therefore,

M=(−1+2, 3+3, 1−1)=(1,6,0).M=(-1+2,\ 3+3,\ 1-1)=(1,6,0).M=(−1+2, 3+3, 1−1)=(1,6,0).
  1. Reflect PPP in the line

Since MMM is midpoint of PPP and its image QQQ,

Q=2M−P.Q=2M-P.Q=2M−P.

So

Q=(2⋅1−5, 2⋅6−4, 2⋅0−2)=(−3,8,−2).Q=(2\cdot1-5,\ 2\cdot6-4,\ 2\cdot0-2)=(-3,8,-2).Q=(2⋅1−5, 2⋅6−4, 2⋅0−2)=(−3,8,−2).

Hence,

α1=−3,α2=8,α3=−2.\alpha_1=-3,\quad \alpha_2=8,\quad \alpha_3=-2.α1​=−3,α2​=8,α3​=−2.

Then

∑i=13αi=−3+8−2=3.\sum_{i=1}^3 \alpha_i=-3+8-2=3.i=1∑3​αi​=−3+8−2=3.

Therefore,

a+∑i=13αi=5+3=8.a+\sum_{i=1}^3 \alpha_i=5+3=8.a+i=1∑3​αi​=5+3=8.
  1. Check options

The value is 888 which corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B
Derived answer: B
So they agree.

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