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Medium Three Dimensional Geometry MCQs for JEE

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Q1MathsUnit 11: Three Dimensional Geometry
The plane x−2y+z−6=0\boldsymbol{x}-\mathbf{2} \boldsymbol{y}+\boldsymbol{z}-\boldsymbol{6}=\mathbf{0} and the line x1=y2=z3\frac{x}{1}=\frac{y}{2}=\frac{z}{3} are related as
Q2MathsUnit 11: Three Dimensional Geometry
Statement-1: The point A(3,1,6)\boldsymbol{A}(\mathbf{3}, \mathbf{1}, \boldsymbol{6}) is the mirror image of the point B(1,3,4)B(1,3,4) in the plane x−y+z=5\boldsymbol{x}-\boldsymbol{y}+\boldsymbol{z}=\mathbf{5} Statement-2: The plane x−y+z=5\boldsymbol{x}-\boldsymbol{y}+\boldsymbol{z}=\mathbf{5} bisects the line segment joining A(3,1,6)\boldsymbol{A}(\boldsymbol{3}, \mathbf{1}, \boldsymbol{6}) and B(1,3,4)\boldsymbol{B}(\mathbf{1}, \boldsymbol{3}, \boldsymbol{4})
Q3MathsUnit 11: Three Dimensional Geometry
How many cubes each of surface area 24cm224 c m^{2} can be made out of a cube of edge measure 1m?1 \mathrm{m} ?
Q4MathsUnit 11: Three Dimensional Geometry
In three dimensions, the coordinate axes of a rectangular cartesian coordinate system are
Q5MathsUnit 11: Three Dimensional Geometry
The equation of the plane passing through the intersection of the planes x+y+z=6\boldsymbol{x}+\boldsymbol{y}+\boldsymbol{z}=\boldsymbol{6} and 2x+3y+4z+5=\boldsymbol{2} \boldsymbol{x}+\boldsymbol{3} \boldsymbol{y}+\boldsymbol{4} \boldsymbol{z}+\boldsymbol{5}= 0,0, and the point (1,1,1) is
Q6MathsUnit 11: Three Dimensional Geometry
Find the equation of the plane through the points A(2,2−1),B(3,4,2)\boldsymbol{A}(\mathbf{2}, \mathbf{2}-\mathbf{1}), \boldsymbol{B}(\mathbf{3}, \mathbf{4}, \mathbf{2}) and C(7,0,6)\boldsymbol{C}(\boldsymbol{7}, \boldsymbol{0}, \boldsymbol{6})
Q7MathsUnit 11: Three Dimensional Geometry
x−21=y−31=z−4−1&x−1k=\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-1} \& \frac{x-1}{k}= y−42=z−52\frac{\boldsymbol{y}-\boldsymbol{4}}{\boldsymbol{2}}=\frac{\boldsymbol{z}-\boldsymbol{5}}{\boldsymbol{2}} are coplanar then k=?\mathbf{k}=?
Q8MathsUnit 11: Three Dimensional Geometry
The shortest distance between the lines x−54=y−7−5=z+3−5\frac{\boldsymbol{x}-\mathbf{5}}{\mathbf{4}}=\frac{\boldsymbol{y}-\mathbf{7}}{-\mathbf{5}}=\frac{\boldsymbol{z}+\mathbf{3}}{-\mathbf{5}} and x−84=\frac{\boldsymbol{x}-\mathbf{8}}{\mathbf{4}}= y−7−5=z−5−5\frac{y-7}{-5}=\frac{z-5}{-5} is

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