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Mathematics Question on Vector Algebra

Three lines L1:r⃗=λi^, λ ∈ RL_{1} : \vec{r} = \lambda\hat{i},\, \lambda\,\in\,R L2:r⃗=k^+μj^, μ ∈ RL_{2} : \vec{r} = \hat{k} + \mu\hat{j},\,\mu\,\in\,R and L3:r⃗=i^+j^+vk^, v ∈ RL_{3} : \vec{r} = \hat{i} + \hat{j} + v\hat{k}, \,v \,\in\,R are given. For which point(s) QQ on L2L_{2} can we find a point P on L1L_{1} and a point R on L3L_{3} so that P, Q and R are collinear?

A

k^−12j^\hat{k} - \frac{1}{2}\hat{j}

B

k^\hat{k}

C

k^+12j^\hat{k} + \frac{1}{2}\hat{j}

D

k^+j^\hat{k} + \hat{j}

Answer

k^+12j^\hat{k} + \frac{1}{2}\hat{j}

Explanation

Solution

Let P(λ,0,0),Q(0,μ,1),R(1,1,v)P\left(\lambda, 0, 0\right), Q\left(0, \mu, 1\right), R\left(1, 1, v\right) be points. L1,L2L_1, L_2 and L3L_3 respectively
Since P,Q,RP, Q, R are collinear, PQ→\overrightarrow{PQ} iscollinear with QR→\overrightarrow{QR}
Hence =−λ1=μ1−μ=1v−1=\frac{-\lambda}{1}=\frac{\mu}{1-\mu}=\frac{1}{v-1}
For every \mu\,\in\,R-\left\\{0, 1\right\\} there exist unique λ,v ∈ R\lambda, v\,\in\,R
Hence Q cannot have coordinates (0,1,1)(0, 1, 1) and (0,0,1).(0, 0, 1).