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3D Geometry question

2020 · 5 Sep · Shift 1 · Q36
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  5. /2020 · 5 Sep · Shift 1 · Q36

3D Geometry question

2020 · 5 Sep · Shift 1 · Q36

JEE MainMathematics3D GeometryMCQ+4 / −1
If (a, b, c) is the image of the point (1, 2, -3) in the line x+12=y−3−2=z−1{{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}2x+1​=−2y−3​=−1z​, then a + b + c is :
  1. A
    1
  2. B
    2
  3. C
    3
  4. D
    -1
View written solutionFree

Correct answer: B

  1. Interpret the question

The phrase “image of the point (1,2,−3)(1,2,-3)(1,2,−3) in the line” means the reflection of the point across the given line.

The line is

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

So its parametric form is

x=−1+2t,y=3−2t,z=−tx=-1+2t,\qquad y=3-2t,\qquad z=-tx=−1+2t,y=3−2t,z=−t

A point on the line is A(−1,3,0)A(-1,3,0)A(−1,3,0) and a direction vector is d⃗=(2,−2,−1).\vec d=(2,-2,-1).d=(2,−2,−1).

Let the given point be P(1,2,−3).P(1,2,-3).P(1,2,−3). If QQQ is the foot of the perpendicular from PPP to the line, then the reflected point P′(a,b,c)P'(a,b,c)P′(a,b,c) satisfies Q=midpoint of PP′.Q=\text{midpoint of }PP'.Q=midpoint of PP′.

So we first find QQQ.

  1. Find the foot of perpendicular from PPP to the line

A general point on the line is Q(−1+2t, 3−2t, −t).Q(-1+2t,\,3-2t,\,-t).Q(−1+2t,3−2t,−t). Then PQ⃗=Q−P=(−1+2t−1, 3−2t−2, −t+3)=(2t−2, 1−2t, 3−t).\vec{PQ}=Q-P=(-1+2t-1,\,3-2t-2,\,-t+3)=(2t-2,\,1-2t,\,3-t).PQ​=Q−P=(−1+2t−1,3−2t−2,−t+3)=(2t−2,1−2t,3−t).

Since QQQ is the foot of the perpendicular, we must have PQ⃗⋅d⃗=0.\vec{PQ}\cdot \vec d=0.PQ​⋅d=0. Thus,

(2t−2, 1−2t, 3−t)⋅(2,−2,−1)=0(2t-2,\,1-2t,\,3-t)\cdot(2,-2,-1)=0(2t−2,1−2t,3−t)⋅(2,−2,−1)=0 2(2t−2)+(−2)(1−2t)+(−1)(3−t)=02(2t-2)+(-2)(1-2t)+(-1)(3-t)=02(2t−2)+(−2)(1−2t)+(−1)(3−t)=0 4t−4−2+4t−3+t=04t-4-2+4t-3+t=04t−4−2+4t−3+t=0 9t−9=09t-9=09t−9=0 t=1.t=1.t=1.

Therefore, Q=(−1+2, 3−2, −1)=(1,1,−1).Q=(-1+2,\,3-2,\,-1)=(1,1,-1).Q=(−1+2,3−2,−1)=(1,1,−1).

  1. Use midpoint relation for reflection

If P′(a,b,c)P'(a,b,c)P′(a,b,c) is the reflection of P(1,2,−3)P(1,2,-3)P(1,2,−3) in the line, then Q(1,1,−1)Q(1,1,-1)Q(1,1,−1) is the midpoint of PP′PP'PP′. So,

(1+a2,2+b2,−3+c2)=(1,1,−1).\left(\frac{1+a}{2},\frac{2+b}{2},\frac{-3+c}{2}\right)=(1,1,-1).(21+a​,22+b​,2−3+c​)=(1,1,−1).

Equating coordinates:

1+a2=1⇒a=1,\frac{1+a}{2}=1 \Rightarrow a=1,21+a​=1⇒a=1, 2+b2=1⇒b=0,\frac{2+b}{2}=1 \Rightarrow b=0,22+b​=1⇒b=0, −3+c2=−1⇒c=1.\frac{-3+c}{2}=-1 \Rightarrow c=1.2−3+c​=−1⇒c=1.

Hence, P′=(1,0,1).P'=(1,0,1).P′=(1,0,1).

  1. Compute a+b+ca+b+ca+b+c

a+b+c=1+0+1=2.a+b+c=1+0+1=2.a+b+c=1+0+1=2.

  1. Check with options

The correct option is 2\boxed{2}2​ which is Option B.

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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