JEE Maths PYQ Practice, chapter-wise

JEE Advanced 2012 Paper 1Question 391

The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,-1,4)$ with the plane $5x-4y-z=1$. If $S$ is the foot of the perpendicular drawn from the point $T(2,1,4)$ to $QR$, then the length of the line segment $PS$ is

3D Geometry

$\frac{1}{\sqrt2}$$\sqrt2$$2$$2\sqrt2$

JEE Advanced 2012 Paper 2Question 392

If the straight lines $\frac{x-1}{2} = \frac{y+1}{k} = \frac{z}{2}$ and $\frac{x+1}{5} = \frac{y+1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is(are)

3D Geometry

More than one option may be correct.

$y + 2z = -1$$y + z = -1$$y - z = -1$$y - 2z = -1$

JEE Advanced 2013 Paper 1Question 393

A line $l$ passing through the origin is perpendicular to the lines $l_1:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},\ -\infty<t<\infty$ $l_2:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},\ -\infty<s<\infty$ Then, the coordinate(s) of the point(s) on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$ is (are)

3D Geometry

More than one option may be correct.

$\left(\frac{7}{3},\frac{7}{3},\frac{5}{3}\right)$$(-1,-1,0)$$(1,1,1)$$\left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right)$

JEE Advanced 2013 Paper 1Question 394

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of perpendiculars lie on the line

3D Geometry

$\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$$\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$$\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$$\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

JEE Advanced 2013 Paper 2Question 395

Two lines $L_1: x=5,\ \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha,\ \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)

3D Geometry

More than one option may be correct.

$1$$2$$3$$4$