Browse Questions

Show that the points $A(-2\hat i + 3\hat j + 5\hat k), B(i+2j+3k) \: and \: C(7\hat i - \hat k)$ are collinear.

Toolbox:
• Two or more vectors are said to be collinear,if they are parallel to be same line,irrespective of their magnitudes and direction.
Step 1:
Given :
$\overrightarrow {OA}=-2\hat i+3\hat j+5\hat k$
$\overrightarrow {OB}=\hat i+2\hat j+3\hat k$
$\overrightarrow {OC}=7\hat i-\hat k$
$\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
$\quad\;=(1+2)\hat i+(2-3)\hat j+(3-5)\hat k$
$\quad\;=3\hat i-\hat j-2\hat k$
Step 2:
$\overrightarrow {BC}=\overrightarrow {OC}-\overrightarrow {OB}$
$\quad\;=(7-1)\hat i+(0-2)\hat j+(-1-3)\hat k$
$\quad\;=6\hat i-2\hat j-4\hat k$
Step 3:
$\overrightarrow {AC}=\overrightarrow {OC}-\overrightarrow {OA}$
$\quad\;=(7+2)\hat i+(0-3)\hat j+(-1-5)\hat k$
$\quad\;=9\hat i-3\hat j-6\hat k$
Step 4:
$\mid\overrightarrow {AB}\mid=\sqrt{3^2+(-1)^2+(-2)^2}$
$\quad\quad=\sqrt{14}$
$\mid\overrightarrow {BC}\mid=\sqrt{6^2+(-2)^2+(-4)^2}$
$\quad\quad=2\sqrt{14}$
$\mid\overrightarrow {AC}\mid=\sqrt{9^2+(-3)^2+(-6)^2}$
$\quad\quad=3\sqrt{14}$
$\therefore\mid \overrightarrow{AC}\mid=\mid \overrightarrow{AB}\mid+\mid \overrightarrow{BC}\mid$
Hence the points A,B and C are collinear.