Question:

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 ?

Updated On: May 24, 2024
  • $3(\alpha+\beta)=-101$
  • $3(\beta+\gamma)=-71$
  • $3(\gamma+\alpha)=-86$
  • $3(\alpha+\beta+\gamma)=-121$
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The Correct Option is C

Solution and Explanation

Given :
Equation of the plane :
\(\vec{r}=-(t+p) \hat{ i }+t \hat{ j }+(1+p) \hat{ k }\)
\(\vec{r}=\hat{k}+t(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k})\)
Standard form of Equation of plane :
\([\vec{r}-\hat{k}\ \ \ \ \ \ \ \ \ \ \hat{i}+\hat{j}\ \ \ \ \ \ \ \ \ -\hat{i}+\hat{k}]=0\)
Therefore, x + y + z = 1  ……. (i)
Coordinates of Q and S :
Q = (10, 15, 20)
S = (α, β, γ)
∴ \(⇒\frac{α-10}{1}=\frac{β-15}{1}=\frac{γ-20}{1}\)
\(=\frac{-2(10+15+20-1)}{3}\)
∴ α = 10 = β = -15 γ - 20 = \(-\frac{83}{3}\)
Therefore, the values are as follows : 
\(α=-\frac{58}{3},\ β=-\frac{43}{3},γ=-\frac{83}{3}\)
∴ 3 (α + β) = −101     so, option (A) is correct.
   3(β + γ) =−71          so, option (B) is correct.
   3(γ + α) = −86         so, option (C) is correct.
  3(α+β+γ)=−129       so, option (D) is incorrect.

So, the correct options are (A), (B) and (C).

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