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:

For 2D vectors, you can also use the slope formula \( m = (y_2 - y_1)/(x_2 - x_1) \). Slopes of AB and BC must be equal.
Slope of AB = \( (8 - (-2))/(5 - 55) = 10/-50 = -1/5 \).
Slope of BC = \( (-52 - 8)/(a - 5) = -60/(a - 5) \).
\( -1/5 = -60/(a - 5) \Rightarrow a - 5 = 300 \Rightarrow a = 305 \).

Answer 1

Concept:
• Collinearity of points: Three points A, B, and C are collinear if the vectors \( \vec{AB} \) and \( \vec{BC} \) are parallel.
• Parallel vectors: \( \vec{u} \parallel \vec{v} \Rightarrow \vec{u} = k\vec{v} \), which implies the components are proportional.

Step 1: Calculate the displacement vectors \( \vec{AB} \) and \( \vec{BC} \)
Let \( \vec{OA} = 55\hat{i} - 2\hat{j} \), \( \vec{OB} = 5\hat{i} + 8\hat{j} \), and \( \vec{OC} = a\hat{i} - 52\hat{j} \). \[ \vec{AB} = \vec{OB} - \vec{OA} = (5 - 55)\hat{i} + (8 - (-2))\hat{j} = -50\hat{i} + 10\hat{j} \] \[ \vec{BC} = \vec{OC} - \vec{OB} = (a - 5)\hat{i} + (-52 - 8)\hat{j} = (a - 5)\hat{i} - 60\hat{j} \]

Step 2: Apply the condition of collinearity
Since the toys are on the same straight line, \( \vec{AB} \) and \( \vec{BC} \) must be parallel. The ratio of their \( \hat{j} \)-components is: \[ \frac{-60}{10} = -6 \] For them to be parallel, the ratio of the \( \hat{i} \)-components must be the same: \[ \frac{a - 5}{-50} = -6 \]

Step 3: Solve for 'a'
\[ a - 5 = (-6) \times (-50) \] \[ a - 5 = 300 \] \[ a = 305 \]