To solve the given problem, we analyze the vector equations and the system of linear equations step-by-step:
1. Vector Relations
We are given:
u ⃗ × v ⃗ = w ⃗ and v ⃗ × w ⃗ = u ⃗ \vec{u} \times \vec{v} = \vec{w} \quad \text{and} \quad \vec{v} \times \vec{w} = \vec{u} u × v = w and v × w = u w ⃗ = ı ^ + ȷ ^ − 2 k ^ \vec{w} = \hat{\imath} + \hat{\jmath} - 2\hat{k} w = ^ + ^ − 2 k ^
Taking the magnitude of both sides:
∣ u ⃗ × v ⃗ ∣ = ∣ w ⃗ ∣ ⟹ ∣ u ⃗ ∣ ∣ v ⃗ ∣ sin θ = ∣ w ⃗ ∣ |\vec{u} \times \vec{v}| = |\vec{w}| \implies |\vec{u}| |\vec{v}| \sin\theta = |\vec{w}| ∣ u × v ∣ = ∣ w ∣ ⟹ ∣ u ∣∣ v ∣ sin θ = ∣ w ∣ ∣ v ⃗ × w ⃗ ∣ = ∣ u ⃗ ∣ ⟹ ∣ v ⃗ ∣ ∣ w ⃗ ∣ sin ϕ = ∣ u ⃗ ∣ |\vec{v} \times \vec{w}| = |\vec{u}| \implies |\vec{v}| |\vec{w}| \sin\phi = |\vec{u}| ∣ v × w ∣ = ∣ u ∣ ⟹ ∣ v ∣∣ w ∣ sin ϕ = ∣ u ∣
From w ⃗ = u ⃗ × v ⃗ \vec{w} = \vec{u} \times \vec{v} w = u × v w ⃗ ⊥ u ⃗ \vec{w} \perp \vec{u} w ⊥ u w ⃗ ⊥ v ⃗ \vec{w} \perp \vec{v} w ⊥ v u ⃗ = v ⃗ × w ⃗ \vec{u} = \vec{v} \times \vec{w} u = v × w u ⃗ ⊥ v ⃗ \vec{u} \perp \vec{v} u ⊥ v u ⃗ ⊥ w ⃗ \vec{u} \perp \vec{w} u ⊥ w
Thus, the vectors u ⃗ , v ⃗ , w ⃗ \vec{u}, \vec{v}, \vec{w} u , v , w sin θ = sin ϕ = 1 \sin\theta = \sin\phi = 1 sin θ = sin ϕ = 1 ∣ u ⃗ ∣ ∣ v ⃗ ∣ = ∣ w ⃗ ∣ and ∣ v ⃗ ∣ ∣ w ⃗ ∣ = ∣ u ⃗ ∣ |\vec{u}| |\vec{v}| = |\vec{w}| \quad \text{and} \quad |\vec{v}| |\vec{w}| = |\vec{u}| ∣ u ∣∣ v ∣ = ∣ w ∣ and ∣ v ∣∣ w ∣ = ∣ u ∣
Multiplying these two equations:
∣ u ⃗ ∣ ∣ v ⃗ ∣ 2 ∣ w ⃗ ∣ = ∣ w ⃗ ∣ ∣ u ⃗ ∣ |\vec{u}| |\vec{v}|^2 |\vec{w}| = |\vec{w}| |\vec{u}| ∣ u ∣∣ v ∣ 2 ∣ w ∣ = ∣ w ∣∣ u ∣
Since ∣ w ⃗ ∣ = 1 2 + 1 2 + ( − 2 ) 2 = 6 ≠ 0 |\vec{w}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6} \neq 0 ∣ w ∣ = 1 2 + 1 2 + ( − 2 ) 2 = 6 = 0 ∣ u ⃗ ∣ = ∣ v ⃗ ∣ ∣ w ⃗ ∣ ≠ 0 |\vec{u}| = |\vec{v}||\vec{w}| \neq 0 ∣ u ∣ = ∣ v ∣∣ w ∣ = 0 ∣ v ⃗ ∣ 2 = 1 ⟹ ∣ v ⃗ ∣ = 1 |\vec{v}|^2 = 1 \implies |\vec{v}| = 1 ∣ v ∣ 2 = 1 ⟹ ∣ v ∣ = 1 ∣ u ⃗ ∣ = ∣ w ⃗ ∣ = 6 ⟹ ∣ u ⃗ ∣ 2 = α 2 + β 2 + γ 2 = 6 |\vec{u}| = |\vec{w}| = \sqrt{6} \implies |\vec{u}|^2 = \alpha^2 + \beta^2 + \gamma^2 = 6 ∣ u ∣ = ∣ w ∣ = 6 ⟹ ∣ u ∣ 2 = α 2 + β 2 + γ 2 = 6
Thus, for (P) :
∣ v ⃗ ∣ 2 = 1 ⟹ ( P ) → ( 2 ) |\vec{v}|^2 = 1 \implies (\text{P}) \rightarrow (2) ∣ v ∣ 2 = 1 ⟹ ( P ) → ( 2 )
2. System of Linear Equations
The homogeneous system in α , β , γ \alpha, \beta, \gamma α , β , γ
− t α + β + γ = 0 α − t β + γ = 0 α + β − t γ = 0 \begin{aligned}
-t\alpha + \beta + \gamma &= 0 \\
\alpha - t\beta + \gamma &= 0 \\
\alpha + \beta - t\gamma &= 0
\end{aligned} − t α + β + γ α − tβ + γ α + β − t γ = 0 = 0 = 0 Since u ⃗ ≠ 0 ⃗ \vec{u} \neq \vec{0} u = 0 ( α , β , γ ) ≠ ( 0 , 0 , 0 ) (\alpha, \beta, \gamma) \neq (0,0,0) ( α , β , γ ) = ( 0 , 0 , 0 )
∣ − t 1 1 1 − t 1 1 1 − t ∣ = 0 \begin{vmatrix}
-t & 1 & 1 \\
1 & -t & 1 \\
1 & 1 & -t
\end{vmatrix} = 0 − t 1 1 1 − t 1 1 1 − t = 0 Expanding the determinant:
− t ( t 2 − 1 ) − 1 ( − t − 1 ) + 1 ( 1 + t ) = 0 -t(t^2 - 1) - 1(-t - 1) + 1(1 + t) = 0 − t ( t 2 − 1 ) − 1 ( − t − 1 ) + 1 ( 1 + t ) = 0 − ( t + 1 ) ( t 2 − t − 2 ) = 0 -(t+1)(t^2 - t - 2) = 0 − ( t + 1 ) ( t 2 − t − 2 ) = 0 − ( t + 1 ) 2 ( t − 2 ) = 0 -(t+1)^2(t-2) = 0 − ( t + 1 ) 2 ( t − 2 ) = 0
Hence, the possible values for t t t t = 2 t = 2 t = 2 t = − 1 t = -1 t = − 1
3. Analysis of Cases
Case 1: t = 2 t = 2 t = 2 The equations become:
− 2 α + β + γ = 0 , α − 2 β + γ = 0 , α + β − 2 γ = 0 -2\alpha + \beta + \gamma = 0, \quad \alpha - 2\beta + \gamma = 0, \quad \alpha + \beta - 2\gamma = 0 − 2 α + β + γ = 0 , α − 2 β + γ = 0 , α + β − 2 γ = 0 α = β = γ \alpha = \beta = \gamma α = β = γ
Using ∣ u ⃗ ∣ 2 = α 2 + β 2 + γ 2 = 6 |\vec{u}|^2 = \alpha^2 + \beta^2 + \gamma^2 = 6 ∣ u ∣ 2 = α 2 + β 2 + γ 2 = 6 3 α 2 = 6 ⟹ α 2 = 2 ⟹ α = ± 2 3\alpha^2 = 6 \implies \alpha^2 = 2 \implies \alpha = \pm\sqrt{2} 3 α 2 = 6 ⟹ α 2 = 2 ⟹ α = ± 2
Thus, if α = 2 \alpha = \sqrt{2} α = 2 t = 2 t = 2 t = 2 t + 3 = 2 + 3 = 5 ⟹ ( S ) → ( 5 ) t + 3 = 2 + 3 = 5 \implies (\text{S}) \rightarrow (5) t + 3 = 2 + 3 = 5 ⟹ ( S ) → ( 5 )
Case 2: t = − 1 t = -1 t = − 1 The system reduces to a single equation:
α + β + γ = 0 \alpha + \beta + \gamma = 0 α + β + γ = 0
Since u ⃗ ⋅ w ⃗ = 0 \vec{u} \cdot \vec{w} = 0 u ⋅ w = 0 ( α ı ^ + β ȷ ^ + γ k ^ ) ⋅ ( ı ^ + ȷ ^ − 2 k ^ ) = 0 ⟹ α + β − 2 γ = 0 (\alpha\hat{\imath} + \beta\hat{\jmath} + \gamma\hat{k}) \cdot (\hat{\imath} + \hat{\jmath} - 2\hat{k}) = 0 \implies \alpha + \beta - 2\gamma = 0 ( α ^ + β ^ + γ k ^ ) ⋅ ( ^ + ^ − 2 k ^ ) = 0 ⟹ α + β − 2 γ = 0
Subtracting the two equations:
( α + β + γ ) − ( α + β − 2 γ ) = 0 ⟹ 3 γ = 0 ⟹ γ = 0 (\alpha + \beta + \gamma) - (\alpha + \beta - 2\gamma) = 0 \implies 3\gamma = 0 \implies \gamma = 0 ( α + β + γ ) − ( α + β − 2 γ ) = 0 ⟹ 3 γ = 0 ⟹ γ = 0
Since γ = 0 \gamma = 0 γ = 0 β = − α \beta = -\alpha β = − α ∣ u ⃗ ∣ 2 = α 2 + β 2 + γ 2 = 6 |\vec{u}|^2 = \alpha^2 + \beta^2 + \gamma^2 = 6 ∣ u ∣ 2 = α 2 + β 2 + γ 2 = 6 α 2 + ( − α ) 2 + 0 = 6 ⟹ 2 α 2 = 6 ⟹ α 2 = 3 ⟹ α = ± 3 \alpha^2 + (-\alpha)^2 + 0 = 6 \implies 2\alpha^2 = 6 \implies \alpha^2 = 3 \implies \alpha = \pm\sqrt{3} α 2 + ( − α ) 2 + 0 = 6 ⟹ 2 α 2 = 6 ⟹ α 2 = 3 ⟹ α = ± 3
If α = 3 \alpha = \sqrt{3} α = 3
γ = 0 ⟹ γ 2 = 0 ⟹ ( Q ) → ( 1 ) \gamma = 0 \implies \gamma^2 = 0 \implies (\text{Q}) \rightarrow (1) γ = 0 ⟹ γ 2 = 0 ⟹ ( Q ) → ( 1 ) β = − 3 ⟹ ( β + γ ) 2 = ( − 3 + 0 ) 2 = 3 ⟹ ( R ) → ( 4 ) \beta = -\sqrt{3} \implies (\beta + \gamma)^2 = (-\sqrt{3} + 0)^2 = 3 \implies (\text{R}) \rightarrow (4) β = − 3 ⟹ ( β + γ ) 2 = ( − 3 + 0 ) 2 = 3 ⟹ ( R ) → ( 4 )
Conclusion
Matching all entries:
( P ) → ( 2 ) (\text{P}) \rightarrow (2) ( P ) → ( 2 ) ( Q ) → ( 1 ) (\text{Q}) \rightarrow (1) ( Q ) → ( 1 ) ( R ) → ( 4 ) (\text{R}) \rightarrow (4) ( R ) → ( 4 ) ( S ) → ( 5 ) (\text{S}) \rightarrow (5) ( S ) → ( 5 )
This corresponds to Option A .