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3D Geometry question

2023 · 25 Jan · Shift 2 · Q23
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  5. /2023 · 25 Jan · Shift 2 · Q23

3D Geometry question

2023 · 25 Jan · Shift 2 · Q23

JEE MainMathematics3D GeometryMCQ+4 / −1
The foot of perpendicular of the point (2, 0, 5) on the line x+12=y−15=z+1−1{{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}2x+1​=5y−1​=−1z+1​ is (α,β,γ\alpha,\beta,\gammaα,β,γ). Then, which of the following is NOT correct?
  1. A
    αβ=−8\frac{\alpha}{\beta}=-8βα​=−8
  2. B
    αβγ=415\frac{\alpha \beta}{\gamma}=\frac{4}{15}γαβ​=154​
  3. C
    βγ=−5\frac{\beta}{\gamma}=-5γβ​=−5
  4. D
    γα=58\frac{\gamma}{\alpha}=\frac{5}{8}αγ​=85​
View written solutionFree

Correct answer: C

  1. Write the line in parametric form

Given x+12=y−15=z+1−1=t\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=t2x+1​=5y−1​=−1z+1​=t so x=2t−1,y=5t+1,z=−t−1.x=2t-1,\quad y=5t+1,\quad z=-t-1.x=2t−1,y=5t+1,z=−t−1.

Thus a general point on the line is Q(t)=(2t−1, 5t+1, −t−1).Q(t)=(2t-1,\,5t+1,\,-t-1).Q(t)=(2t−1,5t+1,−t−1).

The given point is P=(2,0,5).P=(2,0,5).P=(2,0,5).

  1. Use perpendicularity condition

The direction vector of the line is d⃗=(2,5,−1).\vec d=(2,5,-1).d=(2,5,−1).

If Q(t)Q(t)Q(t) is the foot of the perpendicular from PPP to the line, then PQ→=Q(t)−P=(2t−3, 5t+1, −t−6)\overrightarrow{PQ}=Q(t)-P=(2t-3,\,5t+1,\,-t-6)PQ​=Q(t)−P=(2t−3,5t+1,−t−6) must be perpendicular to d⃗\vec dd.

So, PQ→⋅d⃗=0.\overrightarrow{PQ}\cdot \vec d=0.PQ​⋅d=0.

Compute: (2t−3)(2)+(5t+1)(5)+(−t−6)(−1)=0(2t-3)(2)+(5t+1)(5)+(-t-6)(-1)=0(2t−3)(2)+(5t+1)(5)+(−t−6)(−1)=0 4t−6+25t+5+t+6=04t-6+25t+5+t+6=04t−6+25t+5+t+6=0 30t+5=030t+5=030t+5=0 t=−16.t=-\frac{1}{6}.t=−61​.

  1. Find the foot of perpendicular

Substitute t=−16t=-\frac16t=−61​ into the parametric equations:

α=x=2(−16)−1=−13−1=−43,\alpha=x=2\left(-\frac16\right)-1=-\frac13-1=-\frac43,α=x=2(−61​)−1=−31​−1=−34​, β=y=5(−16)+1=−56+1=16,\beta=y=5\left(-\frac16\right)+1=-\frac56+1=\frac16,β=y=5(−61​)+1=−65​+1=61​, γ=z=−(−16)−1=16−1=−56.\gamma=z=-\left(-\frac16\right)-1=\frac16-1=-\frac56.γ=z=−(−61​)−1=61​−1=−65​.

Hence, (α,β,γ)=(−43,16,−56).(\alpha,\beta,\gamma)=\left(-\frac43,\frac16,-\frac56\right).(α,β,γ)=(−34​,61​,−65​).

  1. Check each option

Option A

αβ=−4316=−43⋅6=−8.\frac{\alpha}{\beta}=\frac{-\frac43}{\frac16}=-\frac43\cdot 6=-8.βα​=61​−34​​=−34​⋅6=−8. So A is correct.

Option B

αβγ=(−43)(16)−56\frac{\alpha\beta}{\gamma}=\frac{\left(-\frac43\right)\left(\frac16\right)}{-\frac56}γαβ​=−65​(−34​)(61​)​ =−418−56=−29−56=29⋅65=1245=415.=\frac{-\frac{4}{18}}{-\frac56}=\frac{-\frac29}{-\frac56}=\frac29\cdot\frac65=\frac{12}{45}=\frac{4}{15}.=−65​−184​​=−65​−92​​=92​⋅56​=4512​=154​. So B is correct.

Option C

βγ=16−56=16⋅6−5=−15.\frac{\beta}{\gamma}=\frac{\frac16}{-\frac56}=\frac16\cdot\frac6{-5}=-\frac15.γβ​=−65​61​​=61​⋅−56​=−51​. This is not equal to −5-5−5. So C is not correct.

Option D

γα=−56−43=56⋅34=1524=58.\frac{\gamma}{\alpha}=\frac{-\frac56}{-\frac43}=\frac56\cdot\frac34=\frac{15}{24}=\frac58.αγ​=−34​−65​​=65​⋅43​=2415​=85​. So D is correct.

  1. Conclusion

The only option which is NOT correct is C.\boxed{\text{C}}.C​.

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