3D Geometry

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JEE Advanced 2009 Paper 2Question 386

A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $$2x+y+z=9$$ at point $Q$. The length of the line segment $PQ$ equals

3D Geometry

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

JEE Advanced 2010 Paper 1Question 387

Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

3D Geometry

$x+2y-2z=0$$3x+2y-2z=0$$x-2y+z=0$$5x+2y-4z=0$

JEE Advanced 2010 Paper 1Question 388

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt6$, then $|d|$ is

3D Geometry

Your answer

JEE Advanced 2010 Paper 2Question 389

If the distance of the point $P(1,-2,1)$ from the plane $x+2y-2z=\alpha$, where $\alpha>0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

3D Geometry

$\left(\frac{8}{3},\frac{4}{3},-\frac{7}{3}\right)$$\left(\frac{4}{3},-\frac{4}{3},\frac{1}{3}\right)$$\left(\frac{1}{3},\frac{2}{3},\frac{10}{3}\right)$$\left(\frac{2}{3},-\frac{1}{3},\frac{5}{2}\right)$

JEE Advanced 2012 Paper 2Question 390

The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$ is

3D Geometry

$5x - 11y + z = 17$$\sqrt{2}x + y = 3\sqrt{2} - 1$$x + y + z = \sqrt{3}$$x - \sqrt{2}y = 1 - \sqrt{2}$