arthham

JEE Advanced 2023 · Introduction to Three Dimensional Geometry

Let PP be the plane 3x+2y+3z=16\sqrt{3}x + 2y + 3z = 16 and let S={αi^+βj^+γk^:α2+β2+γ2=1S = \{\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} : \alpha^{2} + \beta^{2} + \gamma^{2} = 1 and the distance of (α,β,γ)(\alpha, \beta, \gamma) from the plane PP is 72}\frac{7}{2}\}. Let u⃗,v⃗\vec{u}, \vec{v} and w⃗\vec{w} be three distinct vectors in SS such that ∣u⃗−v⃗∣=∣v⃗−w⃗∣=∣w⃗−u⃗∣|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|. Let VV be the volume of the parallelepiped determined by vectors u⃗,v⃗\vec{u}, \vec{v} and w⃗\vec{w}. Then the value of 803V\frac{80}{\sqrt{3}}V is

Answer: 4545

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Let P be the plane √3x + 2y + 3z = 16 and let S = {αî + βĵ + γk̂ : α² + β² + γ² = 1 and the distance of (α, β, γ) from the plane P is 7/2}. Let u⃗, v⃗ and w⃗ be three distinct vectors in S such that |u⃗ - v⃗| = |v⃗ - w⃗| = |w⃗ - u⃗|. Let V be the volume of the parallelepiped determined by vectors u⃗, v⃗ and w⃗. Then the value of 80/√3V is P: 3x+2y+3z=16P:\ \sqrt{3}x+2y+3z=16 S={αi^+βj^+γk^:α2+β2+γ2=1, d((α,β,γ),P)=72}S= \{\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} : \alpha^{2}+\beta^{2}+\gamma^{2}=1,\ d((\alpha,\beta,\gamma),P)=\tfrac{7}{2}\} ∣u⃗−v⃗∣=∣v⃗−w⃗∣=∣w⃗−u⃗∣|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}| Find 803V,V=[u⃗ v⃗ w⃗]\text{Find } \dfrac{80}{\sqrt{3}}V, \quad V=[\vec{u}\ \vec{v}\ \vec{w}] 1. d(r⃗,P)=72⇒n⃗⋅r⃗1.\ d(\vec{r},P)=\dfrac{7}{2} \Rightarrow \vec{n}\cdot\vec{r} 2. S is a circle: centre, radius2.\ S \text{ is a circle: centre, radius} 3. V=∣u⃗⋅(v⃗×w⃗)∣, find 803V3.\ V=|\vec{u}\cdot(\vec{v}\times\vec{w})|, \ \text{find } \dfrac{80}{\sqrt{3}}V ∣r⃗∣=1: α2+β2+γ2=1|\vec{r}|=1: \ \alpha^{2}+\beta^{2}+\gamma^{2}=1 n⃗=(3,2,3),∣n⃗∣=3+4+9=4\vec{n}=(\sqrt{3},2,3), \quad |\vec{n}|=\sqrt{3+4+9}=4 d=∣n⃗⋅r⃗−16∣4=72d=\dfrac{|\vec{n}\cdot\vec{r}-16|}{4}=\dfrac{7}{2} ∣n⃗⋅r⃗∣≤∣n⃗∣ ∣r⃗∣=4⇒n⃗⋅r⃗−16<0|\vec{n}\cdot\vec{r}|\le |\vec{n}|\,|\vec{r}|=4 \Rightarrow \vec{n}\cdot\vec{r}-16<0 16−n⃗⋅r⃗=14⇒n⃗⋅r⃗=216-\vec{n}\cdot\vec{r}=14 \Rightarrow \vec{n}\cdot\vec{r}=2 S:∣r⃗∣=1, n⃗⋅r⃗=2S: |\vec{r}|=1, \ \vec{n}\cdot\vec{r}=2 OS=2∣n⃗∣2n⃗=(38,14,38),∣OS∣=12O_{S}=\dfrac{2}{|\vec{n}|^{2}}\vec{n}=\left(\dfrac{\sqrt{3}}{8},\dfrac{1}{4},\dfrac{3}{8}\right), \quad |O_{S}|=\dfrac{1}{2} ρ=1−14=32\rho=\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2} u,v,w∈S,∣u−v∣=∣v−w∣=∣w−u∣u,v,w \in S, \quad |u-v|=|v-w|=|w-u| s=ρ3=32⋅3=32s=\rho\sqrt{3}=\dfrac{\sqrt{3}}{2}\cdot\sqrt{3}=\dfrac{3}{2} V=∣u⃗⋅((v⃗−u⃗)×(w⃗−u⃗))∣V=|\vec{u}\cdot((\vec{v}-\vec{u})\times(\vec{w}-\vec{u}))| (v⃗−u⃗)×(w⃗−u⃗)=2(Area) n^(\vec{v}-\vec{u})\times(\vec{w}-\vec{u}) = 2(\text{Area})\,\hat{n} n^⋅u⃗=24=12\hat{n}\cdot\vec{u}=\dfrac{2}{4}=\dfrac{1}{2} V=2(Area)⋅12=AreaV=2(\text{Area})\cdot\dfrac{1}{2}=\text{Area} Area=34s2=34⋅94\text{Area}=\dfrac{\sqrt{3}}{4}s^{2}=\dfrac{\sqrt{3}}{4}\cdot\dfrac{9}{4} V=9316V=\dfrac{9\sqrt{3}}{16} 803V=803⋅9316\dfrac{80}{\sqrt{3}}V=\dfrac{80}{\sqrt{3}}\cdot\dfrac{9\sqrt{3}}{16} 45\boxed{45}

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