Question

The points with position vectors\[10\hat i + 3\hat j\], \[12\hat j\] and \[a\hat i + 11\hat j\] are collinear then “a” equals to?

Answer 1

Hint: The three position vectors will be collinear if the lines formed by them keeping one of them common will be parallel lie parallel to the same lines. Using the condition for parallel lines, we can find the unknown component of the last vector along the x axis.
Condition for lines to be collinear is $ {a_1}\hat i + {b_1}\hat j $ and $ {a_2}\hat i + {b_2}\hat j $ is $ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} $
\[\overrightarrow {AB} = \vec B - \vec A\] where $ \vec A $ and $ \vec B $ are position vectors.

Complete step-by-step answer:
The positions of three vectors are given. Let these vectors be A, B and C respectively. these vectors along their position vectors are represented as:
\[
  \vec A = 10\hat i + 3\hat j \\
  \vec B = 12\hat j \\
  \vec C = a\hat i + 11\hat j \;
 \]
It is given that all the three points are collinear i.e. lie in a straight line only when line AB will be parallel to AC where A, B and C are collinear vectors:
 $ \overrightarrow {AB} \parallel \overrightarrow {AC} $
Finding their values:
 $ \overrightarrow {AB} $ = difference between position vectors of B and A
\[\Rightarrow \overrightarrow {AB} = \vec B - \vec A\]
Substituting the values, we get:
\[
  \Rightarrow \overrightarrow {AB} = 12\hat j - \left( {10\hat i + 3\hat j} \right) \\
  \Rightarrow \overrightarrow {AB} = 12\hat j - 10\hat i - 3\hat j \\
  \Rightarrow \overrightarrow {AB} = - 10\hat i + 9\hat j \;
 \]
 $ \overrightarrow {AC} $ = difference between position vectors of C and A
\[\Rightarrow \overrightarrow {AC} = \vec C - \vec A\]
Substituting the values, we get:
\[
  \Rightarrow \overrightarrow {AC} = a\hat i + 11\hat j - \left( {10\hat i + 3\hat j} \right) \\
  \Rightarrow \overrightarrow {AC} = a\hat i + 11\hat j - 10\hat i - 3\hat j \\
  \Rightarrow \overrightarrow {AC} = \left( {a - 10} \right)\hat i + 8\hat j \\
 \]
If these two lines are parallel, then the ratio of their corresponding coefficients will be equal.
 $ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} $
The parallel lines are:
\[\overrightarrow {AB} = - 10\hat i + 9\hat j\] and \[\overrightarrow {AC} = \left( {a - 10} \right)\hat i + 8\hat j\]
Here, the corresponding coefficients are:
 $
  {a_1} = - 10 \\
  {a_2} = \left( {a - 10} \right) \\
  $ $
  {b_1} = 9 \\
  {b_2} = 8 \;
  $
Substituting these values in the condition for parallel lines, we get:
 $ \dfrac{{ - 10}}{{\left( {a - 10} \right)}} = \dfrac{9}{8} $
Calculating the value of a by cross multiplication:
 $
  \Rightarrow \dfrac{{ - 10}}{{ - \left( {10 - a} \right)}} = \dfrac{9}{8} \\
  \Rightarrow \dfrac{{10}}{{10 - a}} = \dfrac{9}{8} \\
  \Rightarrow 80 = 90 - 9a \\
  \Rightarrow 9a = 10 \\
   \Rightarrow a = \dfrac{{10}}{9} \;
  $
Therefore, the required value of a is equal to $ \dfrac{{10}}{9} $
So, the correct answer is “ $ \dfrac{{10}}{9} $ ”.

Note: We represent vectors with an arrow over their head and when there is cap over them like $ \hat i $ and $ \hat j $ , such vectors are known as unit vectors and represent components of vectors in x and y axis respectively.
The collinear vectors lie on the same line or parallel lines. We preferred their existence on parallel lines because the condition for the same is equivalence of corresponding coefficients and this condition helped us to find one of the components of the given vectors along x axis.