Question

ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, $ 4\widehat{i}+5\widehat{j}-10\widehat{k} $ , $ 2\widehat{i}-3\widehat{j}+4\widehat{k} $ and \[-\widehat{i}+2\widehat{j}+\widehat{k}\]. Find the vector equation of line BD. Also, reduce it to cartesian form.

Answer 1

Hint: We will use the fact that the diagonals of a parallelogram bisect each other to obtain an equation that has the coordinates of point D as unknowns. We will solve this equation to obtain the coordinates of point D. Then we will find the vector equation of line BD using the coordinates of point B and point D. After that, we will convert this equation into the cartesian form.

Complete step by step answer:
We have a parallelogram ABCD. We know that the diagonals of the parallelogram bisect each other. This means that the diagonal AC and the diagonal BD have the same midpoint. We know that the position vector of a midpoint of two points with position vectors \[\overrightarrow{x}\] and $ \overrightarrow{y} $ is given by $ \dfrac{\overrightarrow{x}+\overrightarrow{y}}{2} $ .
We are given the following position vectors, $ \overrightarrow{\text{A}}=4\widehat{i}+5\widehat{j}-10\widehat{k} $ , $ \overrightarrow{\text{B}}=2\widehat{i}-3\widehat{j}+4\widehat{k} $ and \[\overrightarrow{\text{C}}=-\widehat{i}+2\widehat{j}+\widehat{k}\]. Let us assume that $ \overrightarrow{\text{D}}=a\widehat{i}+b\widehat{j}+c\widehat{k} $ .
We know that,
 $ \text{midpoint of AC}=\text{midpoint of BD} $
Therefore, using the formula for position vector of a midpoint, we get the following,
 $ \dfrac{\overrightarrow{\text{A}}+\overrightarrow{\text{C}}}{2}=\dfrac{\overrightarrow{\text{B}}+\overrightarrow{\text{D}}}{2} $
Substituting the position vectors of all the points, we get
\[\begin{align}
  & \dfrac{4\widehat{i}+5\widehat{j}-10\widehat{k}-\widehat{i}+2\widehat{j}+\widehat{k}}{2}=\dfrac{2\widehat{i}-3\widehat{j}+4\widehat{k}+a\widehat{i}+b\widehat{j}+c\widehat{k}}{2} \\
 & \Rightarrow \dfrac{3\widehat{i}+7\widehat{j}-9\widehat{k}}{2}=\dfrac{\left( 2+a \right)\widehat{i}+\left( -3+b \right)\widehat{j}+\left( 4+c \right)\widehat{k}}{2} \\
 & \therefore \dfrac{3}{2}\widehat{i}+\dfrac{7}{2}\widehat{j}-\dfrac{9}{2}\widehat{k}=\dfrac{\left( 2+a \right)}{2}\widehat{i}+\dfrac{\left( -3+b \right)}{2}\widehat{j}+\dfrac{\left( 4+c \right)}{2}\widehat{k} \\
\end{align}\]
Comparing the coefficients, we have the following,
 $ \begin{align}
  & \dfrac{3}{2}=\dfrac{2+a}{2} \\
 & \Rightarrow a+2=3 \\
 & \therefore a=1 \\
\end{align} $
Similarly, we get
 $ \begin{align}
  & \dfrac{7}{2}=\dfrac{-3+b}{2} \\
 & \Rightarrow -3+b=7 \\
 & \therefore b=10 \\
\end{align} $
And also,
 $ \begin{align}
  & -\dfrac{9}{2}=\dfrac{4+c}{2} \\
 & \Rightarrow 4+c=-9 \\
 & \therefore c=-13 \\
\end{align} $
Hence, the position vector of point D is \[\widehat{i}+10\widehat{j}-13\widehat{k}\].
Now, we will find the vector equation of the line BD in the following manner,
 $ \overrightarrow{\text{BD}}=\overrightarrow{\text{B}}+\lambda \left( \overrightarrow{\text{D}}-\overrightarrow{\text{B}} \right) $ , where $ \lambda $ is a parameter.
Substituting the position vectors of point B and D in the above equation, we get
 $ \begin{align}
  & \overrightarrow{\text{BD}}=2\widehat{i}-3\widehat{j}+4\widehat{k}+\lambda \left( \widehat{i}+10\widehat{j}-13\widehat{k}-2\widehat{i}+3\widehat{j}-4\widehat{k} \right) \\
 & \Rightarrow \overrightarrow{\text{BD}}=2\widehat{i}-3\widehat{j}+4\widehat{k}+\lambda \left( -\widehat{i}+13\widehat{j}-17\widehat{k} \right) \\
 & \therefore \overrightarrow{\text{BD}}=\left( 2-\lambda \right)\widehat{i}+\left( -3+13\lambda \right)\widehat{j}+\left( 4-17\lambda \right)\widehat{k} \\
\end{align} $
Next, we have to convert the above vector equation into cartesian form. We will equate the above equation with $ x\widehat{i}+y\widehat{j}+z\widehat{k} $ as follows,
 $ \left( 2-\lambda \right)\widehat{i}+\left( -3+13\lambda \right)\widehat{j}+\left( 4-17\lambda \right)\widehat{k}=x\widehat{i}+y\widehat{j}+z\widehat{k} $
Comparing the coefficients, we get the following
 $ \begin{align}
  & x=2-\lambda \\
 & \therefore \lambda =\dfrac{x-2}{-1} \\
\end{align} $
 $ \begin{align}
  & y=-3+13\lambda \\
 & \therefore \lambda =\dfrac{y+3}{13} \\
\end{align} $
 $ \begin{align}
  & z=4-17\lambda \\
 & \therefore \lambda =\dfrac{z-4}{-17} \\
\end{align} $
Therefore, the cartesian form of the vector equation is the following,
 $ \lambda =\dfrac{x-2}{-1}=\dfrac{y+3}{13}=\dfrac{z-4}{-17} $

Note:
 It is important that we know the formula for finding the vector equation of a line by using position vectors of two points. The conversion from the cartesian equation to the vector equation is the reverse of the process that we used to convert vector equation into a cartesian form. The calculations in such type of questions can be tricky since we are comparing coefficients in more than one place.