Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

If $$\theta $$ is the angle between line and plane then $$\left( {\frac{\pi }{2} - \theta } \right)$$  is the angle between line and normal to plane given by
$$\eqalign{ & \cos \,\left( {\frac{\pi }{2} - \theta } \right) = \frac{{\left( {\hat i + 2\hat j + 2\hat k} \right).\left( {2\hat i - \hat j + \sqrt \lambda \hat k} \right)}}{{3\sqrt {4 + 1 + \lambda } }} \cr & \cos \left( {\frac{\pi }{2} - \theta } \right) = \frac{{2 - 2 + 2\sqrt \lambda }}{{3 \times \sqrt 5 + \lambda }} \cr & \Rightarrow \sin \,\theta = \frac{{2\sqrt \lambda }}{{3\sqrt 5 + \lambda }} = \frac{1}{3} \cr & \Rightarrow 4\lambda = 5 + \lambda \cr & \Rightarrow \lambda = \frac{5}{3} \cr} $$

Let equation of the variable plane be
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1......\left( 1 \right)$$
This meets the coordinate axes at $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$     and $$C\left( {0,\,0,\,c} \right).$$
Let $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   be the centroid of the $$\Delta ABC.$$   Then
$$\eqalign{ & \alpha = \frac{{a + 0 + 0}}{3},\,\beta = \frac{{0 + b + 0}}{3},\,\gamma = \frac{{0 + 0 + c}}{3} \cr & \therefore \,a = 3\alpha ,\,b = 3\beta ,\,c = 3\gamma ......\left( 2 \right) \cr} $$
Plane $$\left( 1 \right)$$ is at constant distance $$3p$$  from the origin, so
$$\eqalign{ & 3p = \frac{{\left| {\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1} \right|}}{{\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2} + {{\left( {\frac{1}{c}} \right)}^2}} }} \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{9{p^2}}}......\left( 3 \right) \cr} $$
From $$\left( 2 \right)$$ and $$\left( 3 \right),$$ we get
$$\eqalign{ & \frac{1}{{9{\alpha ^2}}} + \frac{1}{{9{\beta ^2}}} + \frac{1}{{9{\gamma ^2}}} = \frac{1}{{9{p^2}}} \cr & \Rightarrow {\alpha ^{ - 2}} + {\beta ^{ - 2}} + {\gamma ^{ - 2}} = {p^{ - 2}} \cr} $$
Generalizing $$\alpha ,\,\beta ,\,\gamma ,$$   locus of centroid $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   is $${x^{ - 2}} + {y^{ - 2}} + {z^{ - 2}} = {p^{ - 2}}$$

Let $$a,\,b,\,c$$  be the intercepts when $$Ox,\,Oy,\,Oz$$    are taken as axes, then the equation of the plane is
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1......\left( {\text{i}} \right)$$
Let $$a,\,b,\,c$$  be the intercepts when $$\left( {OX,\,OY,\,OZ} \right)$$    are taken as axes; then in this case equation of the same plane is
$$\frac{X}{a} + \frac{X}{b} + \frac{X}{c} = 1......\left( {\text{ii}} \right)$$
Now, equations $$\left( {\text{i}} \right)$$ and $$\left( {\text{ii}} \right)$$ are equations of the same plane and in both the cases the origin is same.
Hence, length of the perpendicular drawn from the origin to the plane in both the case must be the same
$$\eqalign{ & \therefore \,\frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }} = \frac{1}{{\sqrt {\frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}} }} \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}} \cr & \therefore \,k = 1 \cr} $$