Question

In square $ ABCD $ , side $ AB $ has column vector $ \left( {\begin{array}{*{20}{c}}
  2 \\
  1
\end{array}} \right) $ . Find two possible column vectors for $ \overrightarrow {BC} $ .

Answer 1

Hint: A column vector represents components of the vector written in a single column of a matrix. For a 2D vector two components of the vector are written. In this question we have been given a column vector for a side of a square. We have to find the possible column vectors for the adjacent side $ BC $ . We can use the conditions that the side $ BC $ will have the same length as side $ AB $ , and it will also be perpendicular to the side $ AB $ .

Complete step by step solution:
We have been the column vector of the side $ AB $ of a square $ ABCD $ . The column vector is given as $ \left( {\begin{array}{*{20}{c}}
  2 \\
  1
\end{array}} \right) $ .
We have to find two possible column vectors for the side $ \overrightarrow {BC} $ .
We can convert column vector in component form as,
 $ \overrightarrow {AB} = \left( {\begin{array}{*{20}{c}}
  2 \\
  1
\end{array}} \right) = 2i + j $ , where $ i $ is the unit vector for x-axis and $ j $ is the unit vector for y-axis.
The magnitude or length of side $ AB $ is $ \left| {\overrightarrow {AB} } \right| = \sqrt {{2^2} + {1^2}} = \sqrt {4 + 1} = \sqrt 5 $
And the unit vector of side $ AB $ is $ \dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} = \dfrac{{2i + j}}{{\sqrt 5 }} = \dfrac{2}{{\sqrt 5 }}i + \dfrac{1}{{\sqrt 5 }}j $
We can assume that the point $ A $ lies on the origin. Then we can get the point $ B $ as $ \left( {2,1} \right) $ .
We know that the scalar product of two perpendicular vectors is zero.
Let us assume a unit vector $ \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}i + \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}j $ that is perpendicular to side $ AB $ .
Two such unit vectors may exist which will result in two different squares as shown in the figure below,

Since $ \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}i + \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}j $ is perpendicular to side $ AB $ , we have,
 $
  \left( {\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}i + \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}j} \right).\left( {\dfrac{2}{{\sqrt 5 }}i + \dfrac{1}{{\sqrt 5 }}j} \right) = 0 \\
   \Rightarrow \dfrac{2}{{\sqrt 5 }}\dfrac{x}{{\sqrt {{x^2} + {y^2}} }} + \dfrac{1}{{\sqrt 5 }}\dfrac{y}{{\sqrt {{x^2} + {y^2}} }} = 0 \\
   \Rightarrow 2x = - y \\
   \Rightarrow x = - \dfrac{1}{2}y \;
  $
We can assume $ y = 2 $ . Then $ x = - 1 $ .
We get a unit vector $ \dfrac{{ - 1}}{{\sqrt {{{\left( { - 1} \right)}^2} + {2^2}} }}i + \dfrac{2}{{\sqrt {{{\left( { - 1} \right)}^2} + {2^2}} }}j = - \dfrac{1}{{\sqrt 5 }}i + \dfrac{2}{{\sqrt 5 }}j $ that is perpendicular to side $ AB $ .
One possible vector for $ BC $ is represented by $ B{C_1} $ . The unit vector is $ - \dfrac{1}{{\sqrt 5 }}i + \dfrac{2}{{\sqrt 5 }}j $ and the magnitude is $ \sqrt 5 $ . Thus,
 $ \overrightarrow {B{C_1}} = \sqrt 5 \left( { - \dfrac{1}{{\sqrt 5 }}i + \dfrac{2}{{\sqrt 5 }}j} \right) = - i + 2j $
The column vector for side $ B{C_1} $ is $ \left( {\begin{array}{*{20}{c}}
  { - 1} \\
  2
\end{array}} \right) $ .
Again, we can assume $ y = - 2 $ . Then $ x = 1 $ .
We get another unit vector $ \dfrac{1}{{\sqrt {{{\left( { - 1} \right)}^2} + {2^2}} }}i + \dfrac{{ - 2}}{{\sqrt {{{\left( { - 1} \right)}^2} + {2^2}} }}j = \dfrac{1}{{\sqrt 5 }}i - \dfrac{2}{{\sqrt 5 }}j $ that is perpendicular to side $ AB $ .
Another possible vector for $ BC $ is represented by $ B{C_2} $ . The unit vector is $ \dfrac{1}{{\sqrt 5 }}i - \dfrac{2}{{\sqrt 5 }}j $ and the magnitude is $ \sqrt 5 $ . Thus,
 $ \overrightarrow {B{C_2}} = \sqrt 5 \left( {\dfrac{1}{{\sqrt 5 }}i - \dfrac{2}{{\sqrt 5 }}j} \right) = i - 2j $
The column vector for side $ B{C_1} $ is $ \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 2}
\end{array}} \right) $ .
Hence, two possible column vectors for $ \overrightarrow {BC} $ are $ \left( {\begin{array}{*{20}{c}}
  { - 1} \\
  2
\end{array}} \right) $ and $ \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 2}
\end{array}} \right) $ .

Note: We took squares on both sides of the given side $ AB $ as both are possible. For the side $ BC $ we have to ensure the conditions that it is perpendicular to side $ AB $ and has the same magnitude as side $ AB $ . We can see that in 2D only two values for $ BC $ would be possible. Also, we can find the point $ C $ as $ \overrightarrow C = \overrightarrow {BC} + \overrightarrow B $ .