Question

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

A. $${a^{ - 2}} + {b^{ - 2}} - {c^{ - 2}} + a{'^{ - 2}} + b{'^{ - 2}} - c{'^{ - 2}} = 0$$
B. $${a^{ - 2}} - {b^{ - 2}} - {c^{ - 2}} + a{'^{ - 2}} - b{'^{ - 2}} - c{'^{ - 2}} = 0$$
C. $${a^{ - 2}} + {b^{ - 2}} + {c^{ - 2}} - a{'^{ - 2}} - b{'^{ - 2}} - c{'^{ - 2}} = 0$$  
D. none of these
Answer :   $${a^{ - 2}} + {b^{ - 2}} + {c^{ - 2}} - a{'^{ - 2}} - b{'^{ - 2}} - c{'^{ - 2}} = 0$$
Solution :
The length of the perpendicular to the plane $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$    from the origin $$ = \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}.$$
The length of the perpendicular to the plane $$\frac{{x'}}{{a'}} + \frac{{y'}}{{b'}} + \frac{{z'}}{{c'}} = 1$$    from the origin (the same point) $$ = \frac{1}{{\sqrt {\frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}} }}.$$
These are equal. So $$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}.$$

Releted MCQ Question on
Geometry >> 3D Geometry and Vectors

Releted Question 1

The scalar $$\vec A.\left( {\vec B + \vec C} \right) \times \left( {\vec A + \vec B + \vec C} \right)$$      equals :

A. $$0$$
B. $$\left[ {\vec A\,\vec B\,\vec C} \right] + \left[ {\vec B\,\vec C\,\vec A} \right]$$
C. $$\left[ {\vec A\,\vec B\,\vec C} \right]$$
D. None of these
Releted Question 2

For non-zero vectors $$\vec a,\,\vec b,\,\vec c,\,\left| {\left( {\vec a \times \vec b} \right).\vec c} \right| = \left| {\vec a} \right|\left| {\vec b} \right|\left| {\vec c} \right|$$       holds if and only if -

A. $$\vec a.\vec b = 0,\,\,\,\vec b.\vec c = 0$$
B. $$\vec b.\vec c = 0,\,\,\,\vec c.\vec a = 0$$
C. $$\vec c.\vec a = 0,\,\,\,\vec a.\vec b = 0$$
D. $$\vec a.\vec b = \vec b.\vec c = \vec c.\vec a = 0$$
Releted Question 3

The volume of the parallelepiped whose sides are given by $$\overrightarrow {OA} = 2i - 2j,\,\,\overrightarrow {OB} = i + j - k,\,\,\overrightarrow {OC} = 3i - k,$$         is :

A. $$\frac{4}{{13}}$$
B. $$4$$
C. $$\frac{2}{7}$$
D. none of these
Releted Question 4

The points with position vectors $$60i + 3j,\,\,40i - 8j,\,\,ai - 52j$$      are collinear if :

A. $$a = - 40$$
B. $$a = 40$$
C. $$a = 20$$
D. none of these

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