JEE Main 2025 — Mathematics Vectors & 3D Geometry

$\begin{aligned} & L_1: \overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=(\lambda-1) \hat{\mathrm{i}}+2(\lambda+1) \hat{\mathrm{j}}+(\lambda+1) \hat{\mathrm{k}} \\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=2 \mu \hat{\mathrm{i}}+(1+7 \mu) \hat{\mathrm{j}}+(1+3 \mu) \hat{\mathrm{k}}\end{aligned}$​L1​:r=(−i^+2j^​+k^)+λ(i^+2j^​+k^)⇒r=(λ−1)i^+2(λ+1)j^​+(λ+1)k^L2​:r=(j^​+k^)+μ(2i^+7j^​+3k^)⇒r=2μi^+(1+7μ)j^​+(1+3μ)k^​ For point of intersection equating respective components $\begin{aligned}
& \Rightarrow \lambda-1=2 \mu \
& 2(\lambda+1)=1+7 \mu \
& \lambda+1=1+3 \mu
\end{aligned}$We get$Weget\begin{aligned}
& \Rightarrow \lambda=3 \text { and } \mu=1 \
& \Rightarrow \vec{a}+\vec{b}=3 \hat{i}+9 \hat{j}+4 \hat{k}
\end{aligned}$$\mathrm{L}_3: \overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\alpha(3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$For$For\alpha=2, \overrightarrow{\mathrm{r}}=8 \hat{\mathrm{i}}+26 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}$