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3D Geometry question

2025 · 4 Apr · Shift 1 · Q40
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  5. /2025 · 4 Apr · Shift 1 · Q40

3D Geometry question

2025 · 4 Apr · Shift 1 · Q40

JEE MainMathematics3D GeometryMCQ+4 / −1
Let AAA and BBB be two distinct points on the line L:x−63=y−72=z−7−2L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}L:3x−6​=2y−7​=−2z−7​. Both AAA and BBB are at a distance 2172 \sqrt{17}217​ from the foot of perpendicular drawn from the point (1,2,3)(1,2,3)(1,2,3) on the line LLL. If OOO is the origin, then OA→⋅OB→\overrightarrow{O A} \cdot \overrightarrow{O B}OA⋅OB is equal to
  1. A
    49
  2. B
    21
  3. C
    47
  4. D
    62
View written solutionFree

Correct answer: C

  1. Parametric form of the line

Given

L:x−63=y−72=z−7−2=tL:\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=tL:3x−6​=2y−7​=−2z−7​=t

So a general point on the line is

P(t)=(6+3t, 7+2t, 7−2t).P(t)=(6+3t,\,7+2t,\,7-2t).P(t)=(6+3t,7+2t,7−2t).

The line passes through

A0=(6,7,7)A_0=(6,7,7)A0​=(6,7,7)

and has direction vector

d⃗=(3,2,−2).\vec d=(3,2,-2).d=(3,2,−2).
  1. Find the foot of perpendicular from Q=(1,2,3)Q=(1,2,3)Q=(1,2,3) to the line

Let the foot be F=P(t0)=(6+3t0,7+2t0,7−2t0)F=P(t_0)=(6+3t_0,7+2t_0,7-2t_0)F=P(t0​)=(6+3t0​,7+2t0​,7−2t0​). Then

A0Q→=Q−A0=(1−6,2−7,3−7)=(−5,−5,−4).\overrightarrow{A_0Q}=Q-A_0=(1-6,2-7,3-7)=(-5,-5,-4).A0​Q​=Q−A0​=(1−6,2−7,3−7)=(−5,−5,−4).

For foot of perpendicular,

(QF→)⋅d⃗=0(\overrightarrow{QF})\cdot \vec d=0(QF​)⋅d=0

which is equivalent to

(A0Q→−t0d⃗)⋅d⃗=0.(\overrightarrow{A_0Q}-t_0\vec d)\cdot \vec d=0.(A0​Q​−t0​d)⋅d=0.

So

t0=A0Q→⋅d⃗d⃗⋅d⃗.t_0=\frac{\overrightarrow{A_0Q}\cdot \vec d}{\vec d\cdot \vec d}.t0​=d⋅dA0​Q​⋅d​.

Now,

A0Q→⋅d⃗=(−5)(3)+(−5)(2)+(−4)(−2)=−15−10+8=−17,\overrightarrow{A_0Q}\cdot \vec d=(-5)(3)+(-5)(2)+(-4)(-2)=-15-10+8=-17,A0​Q​⋅d=(−5)(3)+(−5)(2)+(−4)(−2)=−15−10+8=−17, d⃗⋅d⃗=32+22+(−2)2=9+4+4=17.\vec d\cdot \vec d=3^2+2^2+(-2)^2=9+4+4=17.d⋅d=32+22+(−2)2=9+4+4=17.

Hence

t0=−1717=−1.t_0=\frac{-17}{17}=-1.t0​=17−17​=−1.

Therefore the foot is

F=P(−1)=(6−3,7−2,7+2)=(3,5,9).F=P(-1)=(6-3,7-2,7+2)=(3,5,9).F=P(−1)=(6−3,7−2,7+2)=(3,5,9).
  1. Locate points AAA and BBB on the line at distance 2172\sqrt{17}217​ from FFF

Any point on the line can be written as

P(t)=F+λd⃗P(t)=F+\lambda \vec dP(t)=F+λd

for some scalar λ\lambdaλ. Since ∣d⃗∣=17|\vec d|=\sqrt{17}∣d∣=17​, moving by parameter difference λ\lambdaλ changes distance by

∣λ∣17.|\lambda|\sqrt{17}.∣λ∣17​.

We need this distance to be

217.2\sqrt{17}.217​.

So

∣λ∣17=217  ⟹  ∣λ∣=2.|\lambda|\sqrt{17}=2\sqrt{17}\implies |\lambda|=2.∣λ∣17​=217​⟹∣λ∣=2.

Thus the two points are obtained from t=−1±2t=-1\pm 2t=−1±2, i.e.

t1=1,t2=−3.t_1=1,\qquad t_2=-3.t1​=1,t2​=−3.

Hence

A=P(1)=(9,9,5),A=P(1)=(9,9,5),A=P(1)=(9,9,5), B=P(−3)=(−3,1,13).B=P(-3)=(-3,1,13).B=P(−3)=(−3,1,13).
  1. Compute OA→⋅OB→\overrightarrow{OA}\cdot\overrightarrow{OB}OA⋅OB

Since OOO is the origin,

OA→=(9,9,5),OB→=(−3,1,13).\overrightarrow{OA}=(9,9,5),\qquad \overrightarrow{OB}=(-3,1,13).OA=(9,9,5),OB=(−3,1,13).

Therefore

OA→⋅OB→=9(−3)+9(1)+5(13).\overrightarrow{OA}\cdot\overrightarrow{OB}=9(-3)+9(1)+5(13).OA⋅OB=9(−3)+9(1)+5(13).

Compute:

−27+9+65=47.-27+9+65=47.−27+9+65=47.
  1. Check with options
OA→⋅OB→=47\overrightarrow{OA}\cdot\overrightarrow{OB}=47OA⋅OB=47

So the correct option is C.

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