Start free

Mathematics · 3D Geometry

JEE Advanced 2022 — Paper 1 — Question 12

Let SS be the reflection of a point QQ with respect to the plane given by r⃗=−(t+p)i^+tj^+(1+p)k^\vec{r}=-(t+p) \hat{i}+t \hat{j}+(1+p) \hat{k} where tt, pp are real parameters and i^,j^,k^\hat{i}, \hat{j}, \hat{k} are the unit vectors along the three positive coordinate axes. If the position vectors of QQ and SS are 10i^+15j^+20k^10 \hat{i}+15 \hat{j}+20 \hat{k} and αi^+βj^+γk^\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k} respectively, then which of the following is/are TRUE?

  1. Option A:

    3(α+β)=−1013(\alpha+\beta)=-101

    Correct
  2. Option B:

    3(β+γ)=−713(\beta+\gamma)=-71

    Correct
  3. Option C:

    3(γ+α)=−863(\gamma+\alpha)=-86

    Correct
  4. Option D:

    3(α+β+γ)=−1213(\alpha+\beta+\gamma)=-121

Answer: A, B, C

Step-by-step solution

Clearly plane is given by x+y+z=1x+y+z=1 using mirror image formula

⇒α−101=β−151=γ−201=−2(10+15+20−1)12+12+12=−883\Rightarrow \frac{\alpha-10}{1}=\frac{\beta-15}{1}=\frac{\gamma-20}{1}=\frac{-2(10+15+20-1)}{1^{2}+1^{2}+1^{2}}=-\frac{88}{3}

⇒α=−583;β=−433\Rightarrow \alpha=-\frac{58}{3} ; \beta=-\frac{43}{3} and γ=−283\gamma=-\frac{28}{3}

Answer key and solution verified before publishing.

Practise 3D Geometry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Intersection of lines, line & plane.

More 3D Geometry questions from this paper