Question

If A and B are two points with position vectors \[2\overrightarrow{a}-3\overrightarrow{b}\] and \[6\overrightarrow{b}-\overrightarrow{a}\] respectively then,
Write the position vector of a point P which divides the line segment AB internally in the ratio 1:2

Answer 1

Hint:To solve the above question, we use the section formula i.e.,
If P and Q are the two points represented by position vectors \[\overrightarrow{OP}\,\,and\,\,\overrightarrow{OQ}\] respectively then we can divide the line segment joining points P and Q by a third point R in m:n ratio in two ways
\[\begin{align}
  & \bullet Internal\,\,\,\,\,\, \\
 & \And \\
 & \bullet External \\
\end{align}\]
Then the position vector of internally dividing point R is \[\overrightarrow{OR}\,\,=\dfrac{m\overrightarrow{OQ}+n\overrightarrow{OP}}{m+n}\] and position vector of externally dividing point is \[\overrightarrow{OR}\,\,=\dfrac{m\overrightarrow{OQ}-n\overrightarrow{OP}}{m-n}\].

Complete step-by-step solution:
Here given that:
A and B are two points with position vectors \[2\overrightarrow{a}-3\overrightarrow{b}\] and \[6\overrightarrow{b}-\overrightarrow{a}\],
And a point P divides the line segment AB internally in 1:2 ratio.
Then by comparing with section formula of internal division;
Here m:n=1:2 & \[\overrightarrow{OP}\,\,and\,\,\overrightarrow{OQ}\] are \[2\overrightarrow{a}-3\overrightarrow{b}\] and \[6\overrightarrow{b}-\overrightarrow{a}\]
And \[\overrightarrow{OR}\,\,=\dfrac{m\overrightarrow{OQ}+n\overrightarrow{OP}}{m+n}\]=?
Hence, by section formula:
\[\overrightarrow{OP}\,\]=position vector of point P which divides A and B in the ratio 1:2 internally
\[\Rightarrow \overrightarrow{OP}=\dfrac{\left( 1 \right)\left( 6\overrightarrow{b}-\overrightarrow{a} \right)+\left( 2 \right)\left( 2\overrightarrow{a}-3\overrightarrow{b} \right)}{1+2}\]
\[\begin{align}
  & \Rightarrow \overrightarrow{OP}=\dfrac{6\overrightarrow{b}-\overrightarrow{a}+4\overrightarrow{a}-6\overrightarrow{b}}{1+2} \\
 & \Rightarrow \overrightarrow{OP}=\overrightarrow{a} \\
\end{align}\]
\[\Rightarrow \overrightarrow{OP}=\overrightarrow{a}\]
Hence, the position vector of a point P which divides the line segment AB internally in the ratio 1:2 is
\[\overrightarrow{OP}\]= \[\overrightarrow{a}\]

Note: Here one thing to be noted is that section formula for both the normal lines and position vectors is same as both of them represents a point in the plane which can be an advantage for easy comparison and remembrance and the other thing is: For vector-based problems we have to think graphically to solve the problem easily initially we have to practice locating vectors on the graph on paper the by practice we can imagine the vectors on our own.