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Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

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1 Answer

  • Two or more vectors are said to be collinear if their magnitudes are equal or proportional.
Step 1:
Let the position vectors of points $A,B$ and $C$ be
$\overrightarrow{OA}=\hat i+2\hat j+7\hat k$
$\overrightarrow{OB}=2\hat i+6\hat j+3\hat k$
$\overrightarrow{OC}=3\hat i+10\hat j-\hat k$
Step 2:
Now $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\qquad\quad\;\;=(2\hat i+6\hat j+3\hat k)-(\hat i+2\hat j+7\hat k)$
$\qquad\quad\;\;=\hat i+4\hat j-4\hat k$
Step 3:
$\quad\;\;\;=(3\hat i+10\hat j-\hat k)-(2\hat i+6\hat j+3\hat k)$
$\quad\;\;\;=\hat i+4\hat j-4\hat k$
Step 4:
$\quad\;\;\;=(3\hat i+10\hat j-\hat k)-(\hat i+2\hat j+7\hat k)$
$\quad\;\;\;=2\hat i+8\hat j-8\hat k$
$\quad\;\;\;=2(\hat i+4\hat j-4\hat k)$
Step 5:
Step 6:
Step 7:
Step 8:
Thus $AB+BC=AC$
Hence A,B,C are collinear.
answered May 21, 2013 by sreemathi.v

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