Three lines L₁ : r⃗ = λî, λ ∈ R, L₂ : r⃗ = k̂ + μĵ, μ ∈ R and L₃ : r⃗ = î + ĵ + νk̂, ν ∈ R are given. For which point(s) Q on L₂ can we find a point P on L₁ and a point R on L₃ so that P, Q and R are collinear?
L 1 : r ⃗ = λ i ^ , λ ∈ R L_1:\ \vec{r}=\lambda\hat{i}, \quad \lambda\in\mathbb{R} L 1 : r = λ i ^ , λ ∈ R
L 2 : r ⃗ = k ^ + μ j ^ , μ ∈ R L_2:\ \vec{r}=\hat{k}+\mu\hat{j}, \quad \mu\in\mathbb{R} L 2 : r = k ^ + μ j ^ , μ ∈ R
L 3 : r ⃗ = i ^ + j ^ + ν k ^ , ν ∈ R L_3:\ \vec{r}=\hat{i}+\hat{j}+\nu\hat{k}, \quad \nu\in\mathbb{R} L 3 : r = i ^ + j ^ + ν k ^ , ν ∈ R
To find: Q on L 2 such that P , Q , R are collinear \text{To find: } Q \text{ on } L_2 \text{ such that } P,\ Q,\ R \text{ are collinear} To find: Q on L 2 such that P , Q , R are collinear
with P ∈ L 1 and R ∈ L 3 \text{with } P\in L_1 \text{ and } R\in L_3 with P ∈ L 1 and R ∈ L 3
L 1 : ( 0 , 0 , 0 ) , d ⃗ 1 = i ^ L_1:\ (0,0,0), \quad \vec{d}_1=\hat{i} L 1 : ( 0 , 0 , 0 ) , d 1 = i ^
L 2 : ( 0 , 0 , 1 ) , d ⃗ 2 = j ^ L_2:\ (0,0,1), \quad \vec{d}_2=\hat{j} L 2 : ( 0 , 0 , 1 ) , d 2 = j ^
L 3 : ( 1 , 1 , 0 ) , d ⃗ 3 = k ^ L_3:\ (1,1,0), \quad \vec{d}_3=\hat{k} L 3 : ( 1 , 1 , 0 ) , d 3 = k ^
P = ( λ , 0 , 0 ) ∈ L 1 P=(\lambda,0,0)\in L_1 P = ( λ , 0 , 0 ) ∈ L 1
Q = ( 0 , μ , 1 ) ∈ L 2 Q=(0,\mu,1)\in L_2 Q = ( 0 , μ , 1 ) ∈ L 2
R = ( 1 , 1 , ν ) ∈ L 3 R=(1,1,\nu)\in L_3 R = ( 1 , 1 , ν ) ∈ L 3
P , Q , R collinear P,\ Q,\ R \text{ collinear} P , Q , R collinear
⟺ P Q ⃗ ∥ P R ⃗ \Longleftrightarrow \vec{PQ} \parallel \vec{PR} ⟺ P Q ∥ P R
⟺ P Q ⃗ = t P R ⃗ \Longleftrightarrow \vec{PQ}=t\,\vec{PR} ⟺ P Q = t P R
P Q ⃗ = Q − P = ( − λ , μ , 1 ) \vec{PQ}=Q-P=(-\lambda,\ \mu,\ 1) P Q = Q − P = ( − λ , μ , 1 )
P R ⃗ = R − P = ( 1 − λ , 1 , ν ) \vec{PR}=R-P=(1-\lambda,\ 1,\ \nu) P R = R − P = ( 1 − λ , 1 , ν )
( − λ , μ , 1 ) = t ( 1 − λ , 1 , ν ) (-\lambda,\ \mu,\ 1)=t\,(1-\lambda,\ 1,\ \nu) ( − λ , μ , 1 ) = t ( 1 − λ , 1 , ν )
⇒ − λ = t ( 1 − λ ) , μ = t , 1 = t ν \Rightarrow\ -\lambda=t(1-\lambda),\quad \mu=t,\quad 1=t\nu ⇒ − λ = t ( 1 − λ ) , μ = t , 1 = t ν
t = μ , ν = 1 μ t=\mu,\quad \nu=\dfrac{1}{\mu} t = μ , ν = μ 1
λ = μ μ − 1 \lambda=\dfrac{\mu}{\mu-1} λ = μ − 1 μ
μ ≠ 0 and μ ≠ 1 \mu\neq 0 \quad \text{and}\quad \mu\neq 1 μ = 0 and μ = 1
so every Q = ( 0 , μ , 1 ) works except μ = 0 , 1 \text{so every } Q=(0,\mu,1) \text{ works except } \mu=0,1 so every Q = ( 0 , μ , 1 ) works except μ = 0 , 1
Option (2): μ = 0 ⇒ Q = k ^ \text{Option (2): } \mu=0 \Rightarrow Q=\hat{k} Option (2): μ = 0 ⇒ Q = k ^
ν = 1 0 undefined ⇒ rejected \nu=\dfrac{1}{0} \ \text{undefined} \Rightarrow \text{rejected} ν = 0 1 undefined ⇒ rejected
Option (3): μ = 1 ⇒ Q = k ^ + j ^ \text{Option (3): } \mu=1 \Rightarrow Q=\hat{k}+\hat{j} Option (3): μ = 1 ⇒ Q = k ^ + j ^
μ − 1 = 0 ⇒ λ undefined ⇒ rejected \mu-1=0 \Rightarrow \lambda \ \text{undefined} \Rightarrow \text{rejected} μ − 1 = 0 ⇒ λ undefined ⇒ rejected
Option (1): μ = − 1 2 \text{Option (1): } \mu=-\dfrac{1}{2} Option (1): μ = − 2 1
λ = − 1 / 2 − 3 / 2 = 1 3 , ν = − 2 \lambda=\dfrac{-1/2}{-3/2}=\dfrac{1}{3},\quad \nu=-2 λ = − 3/2 − 1/2 = 3 1 , ν = − 2
Option (4): μ = 1 2 \text{Option (4): } \mu=\dfrac{1}{2} Option (4): μ = 2 1
λ = 1 / 2 − 1 / 2 = − 1 , ν = 2 \lambda=\dfrac{1/2}{-1/2}=-1,\quad \nu=2 λ = − 1/2 1/2 = − 1 , ν = 2
P Q ⃗ = ( 1 , 1 2 , 1 ) \vec{PQ}=(1,\ \tfrac{1}{2},\ 1) P Q = ( 1 , 2 1 , 1 )
P R ⃗ = ( 2 , 1 , 2 ) = 2 P Q ⃗ \vec{PR}=(2,\ 1,\ 2)=2\,\vec{PQ} P R = ( 2 , 1 , 2 ) = 2 P Q
∴ Options (1) and (4) \therefore\ \text{Options (1) and (4)} ∴ Options (1) and (4)
Q = k ^ − 1 2 j ^ or Q = k ^ + 1 2 j ^ Q=\hat{k}-\tfrac{1}{2}\hat{j} \quad \text{or}\quad Q=\hat{k}+\tfrac{1}{2}\hat{j} Q = k ^ − 2 1 j ^ or Q = k ^ + 2 1 j ^
