Question

Show that the points With position vectors-2a+3b+5c,a+2b +3c ,7a-c are collinear position vectors _

Answer 1

$\text{Let the given points be A,B and C}$

$\text{Let O be the origin }$

$\text{Then, }$

$\overrightarrow{OA}=-2\overrightarrow{a}+3\overrightarrow{b}+5\overrightarrow{c}$

$\overrightarrow{OB}=\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}$

$\overrightarrow{OC}=7\overrightarrow{a}-\overrightarrow{c}$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$=3\overrightarrow{a}-\overrightarrow{b}-2\overrightarrow{c}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$

$=9\overrightarrow{a}-3\overrightarrow{b}-6\overrightarrow{c}$

$=3(3a\overrightarrow{a}-\overrightarrow{b}-2\overrightarrow{c})$

$=3\overrightarrow{AB}$

$\implies\bf\overrightarrow{AC}=3\overrightarrow{AB}$

$\overrightarrow{AB}\text{ and }\overrightarrow{AC}\text{ are parallel }$

$\text{But A is the common point }$

$\text{Hence, they are collinear }$

Answer 2

Proved below.

Step-by-step explanation:

Given:

Let us assume that O be the origin and P, Q, R be the position vectors of the given points.

$\vec{OP}=-2\vec {a}+3\vec {b}+5\vec {c}$

$\vec {OQ}=\vec {a}+2\vec {b}+3\vec {c}$

$\vec {OR}=7\vec {a}-\vec {c}$

Then $\vec {PQ}=\vec {OQ}-\vec {OP}$

$\vec {PQ}=\vec {a}+2\vec {b}+3\vec {c}-[-2\vec {a}+3\vec {b}+5\vec {c}]$

$\vec {PQ}=\vec {a}+2\vec {b}+3\vec {c}+2\vec {a}-3\vec {b}-5\vec {c}$

$\vec{PQ}=3\vec{a}-\vec{b}-2\vec{c}$                       [1]

$\vec{QR}=\vec{OR}-\vec{OQ}$

$\vec{QR}=7\vec{a}-\vec{c}-[\vec{a}+2\vec{b}+3\vec{c}]$

$\vec{QR}=7\vec{a}-\vec{c}-\vec{a}-2\vec{b}-3\vec{c}$

$\vec{QR}=6\vec{a}-2\vec{b}-4\vec{c}$

$\vec{QR}=2[3\vec{a}-\vec{b}-2\vec{c}]$

$\vec{QR}=2[\vec{PQ}]$                      [from (1)]

Hence the points P, Q, R are collinear.