During a cricket match, the position of the bowler, the wicket keeper and the leg slip fielder are in a line given by \( \vec{B} = 2\hat{i} + \hat{j} \), \( \vec{W} = 6\hat{i} + 12\hat{j} \) and \( \vec{F} = 12\hat{i} + 1\hat{j} \) respectively. Calculate the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder.
Show Hint
To calculate the ratio in which a point divides a line segment, use the section formula:
\[
\vec{P} = \frac{k \vec{Q} + \vec{A}}{k + 1}
\]
where \( k \) is the ratio of division. Equate the components to find \( k \).
We are given the positions of the bowler (\( \vec{B} \)), the wicketkeeper (\( \vec{W} \)), and the leg slip fielder (\( \vec{F} \)) in a straight line. To find the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder, we use the section formula.
Let the ratio be \( k:1 \). Then, the position of the wicketkeeper is given by:
\[
\vec{W} = \frac{k \vec{F} + \vec{B}}{k + 1}
\]
Substitute the values for \( \vec{W}, \vec{B}, \) and \( \vec{F} \):
\[
6\hat{i} + 12\hat{j} = \frac{k(12\hat{i} + 1\hat{j}) + (2\hat{i} + \hat{j})}{k + 1}
\]
Now equating the components:
For \( \hat{i} \)-components:
\[
6 = \frac{k \times 12 + 2}{k + 1} \implies 6(k + 1) = k \times 12 + 2
\]
\[
6k + 6 = 12k + 2 \implies 6k - 12k = 2 - 6 \implies -6k = -4 \implies k = \frac{2}{3}
\]
Thus, the ratio in which the wicketkeeper divides the line segment is \( 2:3 \).
