Vectors & 3D
3D Geometry: JEE Main & Advanced PYQs
461 previous year questions on 3D Geometry from JEE Main and JEE Advanced papers, each with its exam and year, an answer and a solution.
Practice this chapterQuestions 451-461 of 461
- Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then $(\alpha,\beta)$ equals
- Statement - 1 : The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5$. Statement - 2 : The plane $x-y+z=5$ bisects the line segment …
- If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$, then $\lambda$ equals :
- The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line :
- The equation of the plane containing the line $2x-5y+z=3$; $x+y+4z=5$, and parallel to the plane, $x+3y+6z=1$, is :
- Distance between two parallel planes $2x+y+2z=8$ and $4x+2y+4z+5=0$ is :
- The distance of the point $(1,0,2)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16$, is :
- If two lines $L_1$ and $L_2$ in space, are defined by $L_1=\left\{x=\sqrt\lambda\,y+\left(\sqrt\lambda-1\right),\ z=\left(\sqrt\lambda-1\right)y+\sqrt\lambda\right\}$ …
- The direction ratios of normal to the plane through the points $(0,-1,0)$ and $(0,0,1)$ and making an angle $\frac\pi4$ with the plane $y-z+5=0$ are: (i) $2,-1,1$ (ii) …
- The lines $x=ay-1=z-2$ and $x=3y-2=bz-2$, $(ab eq0)$ are coplanar, if :
- Consider the line $L$ given by the equation $\dfrac{x-3}{2}=\dfrac{y-1}{1}=\dfrac{z-2}{1}$. Let $Q$ be the mirror image of the point $(2,3,-1)$ with respect to $L$. Let …
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