Question

Let $${P_r}\left( {{x_r},\,{y_r},\,{z_r}} \right);\,r = 1,\,2,\,3;$$      be three points where $${x_1},\,{x_2},\,{x_3};\,{y_1},\,{y_2},\,{y_3}$$     and $${z_1},\,{z_2},\,{z_3}$$   are each in GP with the same common ratio. Then $${P_1},\,{P_2},\,{P_3}$$   are :

A. coplanar points
B. collinear points  
C. vertices of an equilateral triangle
D. none of these
Answer :   collinear points
Solution :
$${P_1} = \left( {{x_1},\,r{x_1},\,{r^2}{x_1}} \right),\,{P_2} = \left( {{y_1},\,r{y_1},\,{r^2}{y_1}} \right),\,{P_3} = \left( {{z_1},\,r{z_1},\,{r^2}{z_1}} \right)$$
Direction ratios of $${P_1}{P_2}$$  are $${y_1} - {x_1},\,r\left( {{y_1} - {x_1}} \right),\,{r^2}\left( {{y_1} - {x_1}} \right).$$
Direction ratios of $${P_2}{P_3}$$  are $${z_1} - {y_1},\,r\left( {{z_1} - {y_1}} \right),\,{r^2}\left( {{z_1} - {y_1}} \right).$$
$${\text{As }}\frac{{{y_1} - {x_1}}}{{{z_1} - {y_1}}} = \frac{{r\left( {{y_1} - {x_1}} \right)}}{{r\left( {{z_1} - {y_1}} \right)}} = \frac{{{r^2}\left( {{y_1} - {x_1}} \right)}}{{{r^2}\left( {{z_1} - {y_1}} \right)}},\,{P_1}{P_2}||{P_2}{P_3}$$
So $${P_1},\,{P_2},\,{P_3}$$   are collinear.

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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3D Geometry and Vectors


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