Question

The coplanar points $$A,\,B,\,C,\,D$$   are $$\left( {2 - x,\,2,\,2} \right),\,\left( {2,\,2 - y,\,2} \right),\,\left( {2,\,2,\,2 - z} \right)$$         and $$\left( {1,\,1,\,1} \right)$$   respectively. Then :

A. $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$$  
B. $$x + y + z = 1$$
C. $$\frac{1}{{1 - x}} + \frac{1}{{1 - y}} + \frac{1}{{1 - z}} = 1$$
D. none of these
Answer :   $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$$
Solution :
$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \cr & = \left\{ {2\overrightarrow i + \left( {2 - y} \right)\overrightarrow j + 2\overrightarrow k } \right\} - \left\{ {\left( {2 - x} \right)\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k } \right\} \cr & = x\overrightarrow i - y\overrightarrow j \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} \cr & = \left\{ {2\overrightarrow i + 2\overrightarrow j + \left( {2 - z} \right)\overrightarrow k } \right\} - \left\{ {\left( {2 - x} \right)\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k } \right\} \cr & = x\overrightarrow i - z\overrightarrow k \cr & \overrightarrow {AD} = \overrightarrow {OD} - \overrightarrow {OA} \cr & = \left\{ {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right\} - \left\{ {\left( {2 - x} \right)\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k } \right\} \cr & = \left( {x - 1} \right)\overrightarrow i - \overrightarrow j - \overrightarrow k \cr} $$
As these vectors are coplanar, \[\left| \begin{array}{l} \,\,\,\,x\,\,\,\,\,\,\,\,\,\, - y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\, - z\\ x - 1\,\,\,\,\, - 1\,\,\,\,\, - 1 \end{array} \right| = 0\]
On simplification, $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$$

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

Practice More Releted MCQ Question on
3D Geometry and Vectors


Practice More MCQ Question on Maths Section