JEE Main 3D Geometry Practice Questions With Solutions

Consider the line $\mathrm{L}$L passing through the points $(1,2,3)$\left(\right. 1 , 2 , 3 \left.\right) and $(2,3,5)$\left(\right. 2 , 3 , 5 \left.\right). The distance of the point $(\frac{11}{3},\frac{11}{3},\frac{19}{3})$\left(\right. \frac{11}{3} , \frac{11}{3} , \frac{19}{3} \left.\right) from the line $\mathrm{L}$L along the line $\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}$\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2} is equal to

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5} and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1} is:

Let the line $\mathrm{L}$L intersect the lines $x-2=-y=z-1,2(x+1)=2(y-1)=z+1$x - 2 = - y = z - 1 , 2 \left(\right. x + 1 \left.\right) = 2 \left(\right. y - 1 \left.\right) = z + 1 and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}. Then which of the following points lies on $\mathrm{L}$L ?

If the shortest distance between the lines $\frac{x-\lambda }{2}=\frac{y-4}{3}=\frac{z-3}{4}$\frac{x - \lambda}{2} = \frac{y - 4}{3} = \frac{z - 3}{4} and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8} is $\frac{13}{\sqrt{29}}$\frac{13}{\sqrt{29}}, then a value of $\lambda$\lambda is :

Let $P(x,y,z)$P \left(\right. x , y , z \left.\right) be a point in the first octant, whose projection in the $xy$x y-plane is the point $Q$Q. Let $OP=\gamma$O P = \gamma; the angle between $OQ$O Q and the positive $x$x-axis be $\theta$\theta; and the angle between $OP$O P and the positive $z$z-axis be $\varphi$\phi, where $O$O is the origin. Then the distance of $P$P from the $x$x-axis is

If the shortest distance between the lines

$\begin{array}{ll}{L}_1:\overrightarrow{r}=(2+\lambda )\hat{i}+(1-3\lambda )\hat{j}+(3+4\lambda )\hat{k},& \lambda \in \mathbb{R}\\ L_2:\overrightarrow{r}=2(1+\mu )\hat{i}+3(1+\mu )\hat{j}+(5+\mu )\hat{k},& \mu \in \mathbb{R}\end{array}$L_{1} : \overset{\rightarrow}{r} = \left(\right. 2 + \lambda \left.\right) \hat{i} + \left(\right. 1 - 3 \lambda \left.\right) \hat{j} + \left(\right. 3 + 4 \lambda \left.\right) \hat{k} , & \lambda \in \mathbb{R} \\ L_{2} : \overset{\rightarrow}{r} = 2 \left(\right. 1 + \mu \left.\right) \hat{i} + 3 \left(\right. 1 + \mu \left.\right) \hat{j} + \left(\right. 5 + \mu \left.\right) \hat{k} , & \mu \in \mathbb{R}

is $\frac{m}{\sqrt{n}}$\frac{m}{\sqrt{n}}, where $\gcd (m,n)=1$gcd \left(\right. m , n \left.\right) = 1, then the value of $m+n$m + n equals

Let $\mathrm{P}(\alpha ,\beta ,\gamma )$P \left(\right. \alpha , \beta , \gamma \left.\right) be the image of the point $\mathrm{Q}(3,-3,1)$Q \left(\right. 3 , - 3 , 1 \left.\right) in the line $\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1} and $\mathrm{R}$R be the point $(2,5,-1)$\left(\right. 2 , 5 , - 1 \left.\right). If the area of the triangle $\mathrm{PQR}$PQR is $\lambda$\lambda and ${\lambda }^2=14\mathrm{K}$\lambda^{2} = 14 \textrm{ } K, then $\mathrm{K}$K is equal to :

If $A(3,1,-1),B(\frac{5}{3},\frac{7}{3},\frac{1}{3}),C(2,2,1)$A \left(\right. 3 , 1 , - 1 \left.\right) , B \left(\right. \frac{5}{3} , \frac{7}{3} , \frac{1}{3} \left.\right) , C \left(\right. 2 , 2 , 1 \left.\right) and $D(\frac{10}{3},\frac{2}{3},\frac{-1}{3})$D \left(\right. \frac{10}{3} , \frac{2}{3} , \frac{- 1}{3} \left.\right) are the vertices of a quadrilateral $ABCD$A B C D, then its area is

The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5} and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3} is

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