3D Geometry

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JEE Main 2024 (Online) 5th April Evening ShiftQuestion 426

Let the point $(-1,\alpha,\beta)$ lie on the line of the shortest distance between the lines $\dfrac{x+2}{-3}=\dfrac{y-2}{4}=\dfrac{z-5}{2}$ and $\dfrac{x+2}{-1}=\dfrac{y+6}{2}=\dfrac{z-1}{0}$. Then $(\alpha-\beta)^{2}$ is equal to $\_\_\_\_\_\_\_\_$ .

3D Geometry

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JEE Main 2024 (Online) 5th April Evening ShiftQuestion 427

Let $(\alpha,\beta,\gamma)$ be the image of the point $(8,5,7)$ in the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{5}$. Then $\alpha+\beta+\gamma$ is equal to :

3D Geometry

$20$$18$$16$$14$

JEE Main 2020 (Online) 5th September Evening ShiftQuestion 428

If for some $\alpha\in\mathbf{R}$, the lines $L_1:\dfrac{x+1}{2}=\dfrac{y-2}{-1}=\dfrac{z-1}{1}$ and $L_2:\dfrac{x+2}{\alpha}=\dfrac{y+1}{5-\alpha}=\dfrac{z+1}{1}$ are coplanar, then the line $L_2$ passes through the point :

3D Geometry

$(10,-2,-2)$$(10,2,2)$$(2,-10,-2)$$(-2,10,2)$

JEE Main 2020 (Online) 6th September Morning ShiftQuestion 429

The shortest distance between the lines $\dfrac{x-1}{0}=\dfrac{y+1}{-1}=\dfrac{z}{1}$ and $x+y+z+1=0$, $2x-y+z+3=0$ is :

3D Geometry

$1$$\dfrac{1}{\sqrt{2}}$$\dfrac{1}{\sqrt{3}}$$\dfrac{1}{2}$

JEE Main 2025 (Online) 22nd January Morning ShiftQuestion 430

Let $L_1:\dfrac{x-1}{3}=\dfrac{y-1}{-1}=\dfrac{z+1}{0}$ and $L_2:\dfrac{x-2}{2}=\dfrac{y}{0}=\dfrac{z+4}{\alpha}$, $\alpha\in\mathbf{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26\alpha(PB)^{2}$ is $\_\_\_\_\_\_\_\_$ .

3D Geometry

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