3D Geometry

The distance of the point \((1,-5,9)\) from the plane \(x-y+z=5\) measured along the line \(x=y=z\) is :

If the line, \({{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z + 4} \over 3}\,\) lies in the planes, \(lx+my-z=9,\) then \({l^2} + {m^2}\) is equal to :

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 :

The distance of the point \((1, 0, 2)\) from the point of intersection of the line \({{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}\) and the plane \(x - y + z = 16,\) is :

The angle between the lines whose direction cosines satisfy the equations \(l+m+n=0\) and \({l^2} = {m^2} + {n^2}\) is :

The image of the line \({{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\,\) in the plane \(2x-y+z+3=0\) is the line :

If the lines \({{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}\) and \({{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}\) are coplanar, then \(k\) can have :

Distance between two parallel planes \(2x+y+2z=8\) and \(4x+2y+4z+5=0\) is :

A equation of a plane parallel to the plane \(x-2y+2z-5=0\) and at a unit distance from the origin is :

If the line \({{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}\) and \({{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}\) intersect, then \(k\) is equal to :

Statement - 1 : The point \(A(1,0,7)\) is the mirror image of the point
\(B(1,6,3)\) in the line : \({x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}\)
Statement - 2 : The line \({x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}\) bi...

If the angle between the line \(x = {{y - 1} \over 2} = {{z - 3} \over \lambda }\) and the plane
\(x+2y+3z=4\) is \({\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right),\) then \(\lambda\) 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 joining \(A(3, 1, 6)\) and \(B(1, 3, 4).\)

A line \(AB\) in three-dimensional space makes angles \({45^ \circ }\) and \({120^ \circ }\) with the positive \(x\)-axis and the positive \(y\)-axis respectively. If \(AB\) makes an acute angle \(\theta\) with the positive \(z\)-axis, th...

The projections of a vector on the three coordinate axis are \(6,-3,2\) respectively. The direction cosines of the vector are :

Let the line \(\,\,\,\,\,\) \({{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}\) lie in the plane \(\,\,\,\,\,\) \(x + 3y - \alpha z + \beta = 0.\) Then \(\left( {\alpha ,\beta } \right)\) equals

If the straight lines \(\,\,\,\,\,\) \(\,\,\,\,\,\) \({{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}\) \(\,\,\,\,\,\) and\(\,\,\,\,\,\) \({{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}\) intersects at a point, then th...

The line passing through the points \((5,1,a)\) and \((3, b, 1)\) crosses the \(yz\)-plane at the point \(\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)\) . Then

If a line makes an angle of \(\pi /4\) with the positive directions of each of \(x\)-axis and \(y\)-axis, then the angle that the line makes with the positive direction of the \(z\)-axis is :

If \((2,3,5)\) is one end of a diameter of the sphere \({x^2} + {y^2} + {z^2} - 6x - 12y - 2z + 20 = 0,\) then the coordinates of the other end of the diameter are

Let \(L\) be the line of intersection of the planes \(2x+3y+z=1\) and \(x+3y+2z=2.\) If \(L\) makes an angle \(\alpha\) with the positive \(x\)-axis, then cos \(\alpha\) equals

The two lines \(x=ay+b, z=cy+d;\) and \(x=a'y+b' ,\) \(z=c'y+d'\) are perpendicular to each other if :

The image of the point \((-1, 3,4)\) in the plane \(x-2y=0\) is :

The angle between the lines \(2x=3y=-z\) and \(6x=-y=-4z\) is :

If the plane \(2ax-3ay+4az+6=0\) passes through the midpoint of the line joining the centres of the spheres
\({x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13\) and
\({x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8\) then a equals :

The plane \(x+2y-z=4\) cuts the sphere \({x^2} + {y^2} + {z^2} - x + z - 2 = 0\) in a circle of radius

The distance between the line
\(\overrightarrow r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda \left( {i - j + 4k} \right),\) and the plane
\(\overrightarrow r .\left( {\widehat i + 5\widehat j + \widehat k} \right) = 5\) is

If the angel \(\theta\) between the line \({{x + 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}\) and
the plane \(2x - y + \sqrt \lambda \,\,z + 4 = 0\) is such that \(\sin \,\,\theta = {1 \over 3}\) then value of \(\lambda\) is :

The intersection of the spheres
\({x^2} + {y^2} + {z^2} + 7x - 2y - z = 13\) and
\({x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8\)
is the same as the intersection of one of the sphere and the plane

Distance between two parallel planes
\(\,2x + y + 2z = 8\) and \(4x + 2y + 4z + 5 = 0\) is :

A line makes the same angle \(\theta\), with each of the \(x\) and \(z\) axis.
If the angle \(\beta \,\), which it makes with y-axis, is such that \(\,{\sin ^2}\beta = 3{\sin ^2}\theta ,\) then \({\cos ^2}\theta\) equals :

A line with direction cosines proportional to \(2,1,2\) meets each of the lines \(x=y+a=z\) and \(x+a=2y=2z\) . The co-ordinates of each of the points of intersection are given by :

If the straight lines
\(x=1+s,y=-3\)\(- \lambda s,\) \(z = 1 + \lambda s\) and \(x = {t \over 2},y = 1 + t,z = 2 - t,\) with parameters \(s\) and \(t\) respectively, are co-planar, then \(\lambda\) equals :

The shortest distance from the plane \(12x+4y+3z=327\) to the sphere \({x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155\) is

The lines \({{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}\) and \({{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}\) are coplanar if :

The two lines \(x=ay+b,z=cy+d\) and \(x = a'y + b',z = c'y + d'\) will be perpendicular, if and only if :

The radius of the circle in which the sphere
\({x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0\) is cut by the plane
\(x+2y+2z+7=0\) is

Two systems of rectangular axes have the same origin. If a plane cuts then at distances \(a,b,c\) and \(a', b', c'\) from the origin then

A plane which passes through the point \((3,2,0)\) and the line
\({{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}\) is :

The \(d.r.\) of normal to the plane through \((1, 0, 0), (0, 1, 0)\) which makes an angle \(\pi /4\) with plane \(x+y=3\) are :