Question

Let three toys A, B and C be placed in the same straight line. If the position vectors of A, B and C are \( 55\hat{i} - 2\hat{j} \), \( 5\hat{i} + 8\hat{j} \) and \( a\hat{i} - 52\hat{j} \) respectively, find the value of ‘a’.

Hint:

Collinearity for 2D vectors can also be solved using slopes: \( m_{AB} = m_{BC} \). Here, \( \frac{8 - (-2)}{5 - 55} = \frac{-52 - 8}{a - 5} \).

Answer 1

Concept:
• If three points A, B, and C are collinear, then the vectors \( \vec{AB} \) and \( \vec{BC} \) are parallel.
• Parallel vectors have proportional components: if \( \vec{u} = x_1\hat{i} + y_1\hat{j} \) and \( \vec{v} = x_2\hat{i} + y_2\hat{j} \) are parallel, then \( \frac{x_1}{x_2} = \frac{y_1}{y_2} \).

Step 1: Find the vectors \( \vec{AB} \) and \( \vec{BC} \)
\[ \vec{AB} = \vec{B} - \vec{A} = (5 - 55)\hat{i} + (8 - (-2))\hat{j} = -50\hat{i} + 10\hat{j} \]
\[ \vec{BC} = \vec{C} - \vec{B} = (a - 5)\hat{i} + (-52 - 8)\hat{j} = (a - 5)\hat{i} - 60\hat{j} \]

Step 2: Set up the proportionality equation for collinearity
Since the toys are in a straight line:
\[ \frac{-50}{a - 5} = \frac{10}{-60} \]
Simplify the right side:
\[ \frac{-50}{a - 5} = -\frac{1}{6} \]

Step 3: Solve for \( a \)
Cross-multiply the terms:
\[ -50 \times (-6) = 1 \times (a - 5) \]
\[ 300 = a - 5 \]
\[ a = 305 \]