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QJEEAdv22Ma83b2: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma83b2 Solution: Let be the reflection of a point with respect to the plane…

Let SS be the reflection of a point QQ with respect to the plane given by

r⃗=−(t+p)ı^+tȷ^+(1+p)k^\vec{r}=-(t+p) \hat{\imath}+t \hat{\jmath}+(1+p) \hat{k}

where t,pt, p are real parameters and ı^,ȷ^,k^\hat{\imath}, \hat{\jmath}, \hat{k} are the unit vectors along the three positive coordinate axes. If the position vectors of QQ and SS are 10ı^+15ȷ^+20k^10 \hat{\imath}+15 \hat{\jmath}+20 \hat{k} and αı^+βȷ^+γk^\alpha \hat{\imath}+\beta \hat{\jmath}+\gamma \hat{k} respectively, then which of the following is/are TRUE ?

Multiple Choice Options

  • A 3(α+β)=−1013(\alpha+\beta)=-101
  • B 3(β+γ)=−713(\beta+\gamma)=-71
  • C 3(γ+α)=−863(\gamma+\alpha)=-86
  • D 3(α+β+γ)=−1213(\alpha+\beta+\gamma)=-121

Step-by-Step Verified Solution

Given : equation of plane

r⃗=−(t+p)i^+tj^+(1+p)k^\vec{r}=-(t+p) \hat{i}+t \hat{j}+(1+p) \hat{k}

on rearranging we get,

r⃗=k+t+(−i^+j^)+p(−i^+k^)\vec{r}=k+t+(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k})

So, equation of plane in standard form is given by

amp;[r⃗−k^        −i^+j^        −i^+k^]=0amp;⇒[xamp;yamp;z−1−1amp;1amp;0−1amp;0amp;1]=0amp;⇒x+y+z=1.........(1)\begin{aligned}& {[\vec{r}-\hat{k}\,\,\,\,\,\,\,\,-\hat{i}+\hat{j}\,\,\,\,\,\,\,\,-\hat{i}+\hat{k}]=0} \\\\& \Rightarrow\left[\begin{array}{rrr}x & y & z-1 \\-1 & 1 & 0 \\-1 & 0 & 1\end{array}\right]=0 \\\\& \Rightarrow x+y+z=1 .........(1)\end{aligned}

Now given co-ordinate of Q=(10,15,20)Q=(10,15,20)

And Co-ordinates of S=(α,β,γ)S=(\alpha, \beta, \gamma)

Now using the image formula of point and plane we get,

amp;α−101=β−151=γ−201=−2(10+15+20−1)3amp;⇒α−10=β−15=γ−20=−883amp;⇒α=−583,β=−433,γ=−283\begin{aligned}& \frac{\alpha-10}{1}=\frac{\beta-15}{1}=\frac{\gamma-20}{1}=\frac{-2(10+15+20-1)}{3} \\\\& \Rightarrow \alpha-10=\beta-15=\gamma-20=\frac{-88}{3} \\\\& \Rightarrow \alpha=\frac{-58}{3}, \beta=\frac{-43}{3}, \gamma=\frac{-28}{3}\end{aligned}

Now solving all options.

3(α+β)amp;=−1013(β+γ)amp;=−713(γ+α)amp;=−863(α+β+γ)amp;=−129\begin{aligned}3(\alpha+\beta) & =-101 \\\\3(\beta+\gamma) & =-71 \\\\3(\gamma+\alpha) & =-86 \\\\3(\alpha+\beta+\gamma) & =-129\end{aligned}

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