JEE Main 2025 — Mathematics Vectors & 3D Geometry

$$\begin{aligned} & \overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{q}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \mathrm{A} \equiv(1,2,3) B \equiv(\lambda, 4,5) \\ & \text { Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}}|}\right|\end{aligned}$$​p​=2i^+3j^​+4k^,q​=3i^+4j^​+5k^⇒p​×q​=​i^23​j^​34​k^45​​=−i^+2j^​−k^A≡(1,2,3)B≡(λ,4,5) Shortest Distance =​∣P×q​∣AB⋅(P×q​)​​​

$\begin{aligned}
& \frac{1}{\sqrt{6}}=\left|\frac{((\lambda-1) \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-\hat{i}+2 \hat{j}-\hat{k})}{\sqrt{6}}\right| \
& \Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1 \
& \Rightarrow \lambda=3 \pm 1=4,2
\end{aligned}$

Radius of circle passing through points

$\begin{aligned}
& (0,0),(4,2) &(2,4) \
& =\frac{\text { abc }}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll}
1 & 1 & 1 \
0 & 4 & 2 \
0 & 2 & 4
\end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12} \
& =\frac{5 \sqrt{2}}{3}
\end{aligned}$