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JEE Advanced 2022 · Introduction to Three Dimensional Geometry

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^\hat{i}, j^\hat{j}, k^\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 ?

Answers: A. 3(α+β)=−1013(\alpha+\beta)=-101 · B. 3(β+γ)=−713(\beta+\gamma)=-71 · C. 3(γ+α)=−863(\gamma+\alpha)=-86

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r⃗=−(t+p)i^+tj^+(1+p)k^\vec r = -(t+p)\hat{i} + t\hat{j} + (1+p)\hat{k} Q=10i^+15j^+20k^,S=αi^+βj^+γk^Q = 10\hat{i} + 15\hat{j} + 20\hat{k},\quad S = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} Find: which of (1)–(4) is/are TRUE\text{Find: which of (1)--(4) is/are TRUE} x=−(t+p),y=t,z=1+px = -(t+p),\quad y = t,\quad z = 1+p x+y+z=−(t+p)+t+(1+p)=1x + y + z = -(t+p) + t + (1+p) = 1 1. n⃗=(1,1,1), n⃗⋅Q=?1.\ \vec n = (1,1,1),\ \vec n \cdot Q = ? 2. F=foot of ⊥ from Q2.\ F = \text{foot of } \perp \text{ from } Q 3. S=2F−Q3.\ S = 2F - Q n=(1,1,1),d=1n = (1,1,1),\quad d = 1 n⋅Q=10+15+20=45n \cdot Q = 10 + 15 + 20 = 45 F=Q−n⋅Q−d∣n∣2 nF = Q - \dfrac{n \cdot Q - d}{|n|^2}\,n =Q−443(1,1,1)= Q - \dfrac{44}{3}(1,1,1) F=(10−443, 15−443, 20−443)F = \left(10 - \dfrac{44}{3},\ 15 - \dfrac{44}{3},\ 20 - \dfrac{44}{3}\right) F=(−143, 13, 163)F = \left(-\dfrac{14}{3},\ \dfrac13,\ \dfrac{16}{3}\right) S=2F−QS = 2F - Q S=(−583, −433, −283)S = \left(-\dfrac{58}{3},\ -\dfrac{43}{3},\ -\dfrac{28}{3}\right) α=−583,β=−433,γ=−283\alpha = -\dfrac{58}{3},\quad \beta = -\dfrac{43}{3},\quad \gamma = -\dfrac{28}{3} 3(α+β)=3(−583−433)=3(−1013)=−1013(\alpha+\beta) = 3\left(-\dfrac{58}{3}-\dfrac{43}{3}\right) = 3\left(-\dfrac{101}{3}\right) = -101 ⇒(1) TRUE\Rightarrow (1) \text{ TRUE} 3(β+γ)=3(−433−283)=3(−713)=−713(\beta+\gamma) = 3\left(-\dfrac{43}{3}-\dfrac{28}{3}\right) = 3\left(-\dfrac{71}{3}\right) = -71 ⇒(2) TRUE\Rightarrow (2) \text{ TRUE} 3(γ+α)=3(−283−583)=3(−863)=−863(\gamma+\alpha) = 3\left(-\dfrac{28}{3}-\dfrac{58}{3}\right) = 3\left(-\dfrac{86}{3}\right) = -86 ⇒(3) TRUE\Rightarrow (3) \text{ TRUE} 3(α+β+γ)=3(−1293)=−129≠−1213(\alpha+\beta+\gamma) = 3\left(-\dfrac{129}{3}\right) = -129 \neq -121 ⇒(4) FALSE\Rightarrow (4) \text{ FALSE} (1), (2), (3) are TRUE\boxed{(1),\ (2),\ (3) \text{ are TRUE}}

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