3D Geometry

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JEE Main 2024 (Online) 4th April Morning ShiftQuestion 421

If the shortest distance between the lines $\dfrac{x+2}{2}=\dfrac{y+3}{3}=\dfrac{z-5}{4}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{-3}=\dfrac{z+4}{2}$ is $\dfrac{38}{3\sqrt{5}}k$, and $\displaystyle\int_{0}^{k}\left[x^{2}\right]dx=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^{3}$ is equal to $\_\_\_\_\_\_\_\_$ .

3D Geometry

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JEE Main 2024 (Online) 4th April Evening ShiftQuestion 422

Let $P$ be the point of intersection of the lines $\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{1}$ and $\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{2}$. Then, the shortest distance of $P$ from the line $4x=2y=z$ is

3D Geometry

$\dfrac{3\sqrt{14}}{7}$$\dfrac{\sqrt{14}}{7}$$\dfrac{5\sqrt{14}}{7}$$\dfrac{6\sqrt{14}}{7}$

JEE Main 2024 (Online) 4th April Evening ShiftQuestion 423

Consider a line $L$ passing through the points $P(1,2,1)$ and $Q(2,1,-1)$. If the mirror image of the point $A(2,2,2)$ in the line $L$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+6\gamma$ is equal to $\_\_\_\_\_\_\_\_$ .

3D Geometry

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JEE Main 2024 (Online) 5th April Morning ShiftQuestion 424

Let $d$ be the distance of the point of intersection of the lines $\dfrac{x+6}{3}=\dfrac{y}{2}=\dfrac{z+1}{1}$ and $\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}$ from the point $(7,8,9)$. Then $d^{2}+6$ is equal to :

3D Geometry

$69$$72$$75$$78$

JEE Main 2024 (Online) 5th April Morning ShiftQuestion 425

If the line $\dfrac{2-x}{3}=\dfrac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\dfrac{x+3}{3\mu}=\dfrac{1-2y}{6}=\dfrac{5-z}{7}$, then $4\lambda+9\mu$ is equal to :

3D Geometry

$4$$5$$6$$13$