{\frac{2}{3}}=\lambda$$

Any point on $$L_1:(\mu+1, \mu+2,2 \mu+3)$$

Any point on $$L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)$$

JEE Main 2024 (Online) 9th April Evening Shift Mathematics - 3D Geometry Question 83 English Explanation

Now

$$\begin{aligned} & \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \\ & \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \\ & \mu=\frac{2}{3} \\ \end{aligned}$$

$$\begin{aligned} & \text { Point on } L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \\ & d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \\ & d=\sqrt{4+1+4} \\ & d=3 \end{aligned}$$

19:T5c0,

$$L: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu$$

$$L_2$$ के अनुदिश मापा गया: $$L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda$$

$$L_1$$ पर कोई बिंदु: $$(\mu+1, \mu+2,2 \mu+3)$$

$$L_2$$ पर कोई बिंदु $$\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)$$

3d Geometry

Continue this paper

Similar 3d Geometry PYQs

More practice from this question