Vector Algebra
Position Vectors and Section Formula
Grade 12
Question:
<p>If <i>C</i> is the midpoint of <i>AB</i> and <i>P</i> is any point outside <i>AB</i>, then</p>
<p>\(\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}\)</p>
<p>\(\overrightarrow{PA} + \overrightarrow{PB} = \overrightarrow{PC}\)</p>
<p>\(\overrightarrow{PA} + \overrightarrow{PB} + 2\overrightarrow{PC} = 0\)</p>
<p>\(\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = 0\)</p>
Step-by-Step Solution
Key Concept: The position vector of a midpoint is the average of position vectors of endpoints. Any vector equation involving midpoints can be expressed using this fundamental relationship: if C is midpoint of AB, then PC = (PA + PB)/2.
Step 1: Let position vectors of A, B, C, P be a , b , c , p respectively. Step 2: Since C is midpoint of AB: c = ( a + b )/2 Step 3: The vector from P to C is: PC = c - p = ( a + b )/2 - p Step 4: The vectors from P to A and P to B are: PA = a - p and PB = b - p Step 5: Adding them: PA + PB = ( a - p ) + ( b - p ) = a + b - 2 p Step 6: Therefore: PA + PB = 2[( a + b )/2 - p ] = 2 PC Step 7: Rearranging: PC = ( PA + PB )/2 or PA + PB = 2 PC ∴ The fundamental relation is: PA + PB = 2 PC
Correct Answer: A