JEE Advanced 2022 Paper 1 Q12 Mathematics Vectors & 3D Geometry 3D Planes Medium

JEE Advanced 2022 Paper 1 · Q12 · 3D Planes

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

  1. A. $3(\alpha+\beta)=-101$
  2. B. $3(\beta+\gamma)=-71$
  3. C. $3(\gamma+\alpha)=-86$
  4. D. $3(\alpha+\beta+\gamma)=-121$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

The plane: $\vec{r}=t(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k})+\hat{k}$. Direction vectors $\vec{u}=(-1,1,0)$, $\vec{v}=(-1,0,1)$; point on plane $(0,0,1)$. Normal $\vec{n}=\vec{u}\times\vec{v}=(1,1,1)$. Plane equation: $x+y+z=1$. Reflect $Q=(10,15,20)$ across $x+y+z=1$: $Q$ to plane signed distance $=(10+15+20-1)/\sqrt{3}=44/\sqrt{3}$. Image $S=Q-2\cdot\dfrac{44}{3}(1,1,1)=(10-88/3,15-88/3,20-88/3)=(-58/3,-43/3,-28/3)$. So $\alpha=-58/3,\beta=-43/3,\gamma=-28/3$. Check: $3(\alpha+\beta)=-101$ TRUE; $3(\beta+\gamma)=-71$ TRUE; $3(\gamma+\alpha)=-86$ TRUE; $3(\alpha+\beta+\gamma)=-129\ne -121$ FALSE. Answer: A, B, C.

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