Question

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

Answer 1

We know that the position vector of the mid-point of `vec a` and `vec b` is \[\frac{\vec{a} + \vec{b}}{2}\] Let the position vector of point D be  \[x \hat{i} + y \hat{j} + z \hat{k} \] Position vector of mid-point of A and C = Position vector of mid-point of B and D

\[\therefore \frac{\left( 4 \hat{i} + 5 \hat{j} - 10 \hat{k} \right) + \left( - \hat{i} + 2 \hat{j} + \hat{k} \right)}{2} = \frac{\left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \left( x \hat{i}+ y \hat{j} + z \hat{k} \right)}{2}\]

\[ \Rightarrow \frac{3}{2} \hat{i} + \frac{7}{2} \hat{j} - \frac{9}{2} \hat{k}= \left( \frac{x + 2}{2} \right) \hat{i} + \left( \frac{- 3 + y}{2} \right) \hat{j} + \left( \frac{4 + z}{2} \right) \hat{k} \]

\[\text{ Comparing the coefficient of } \hat{i} ,\hat{j}  \text{ and }  \hat{k} , \text{ we get } \]

\[\frac{x + 2}{2} = \frac{3}{2}\]

\[ \Rightarrow x = 1\]

\[\frac{- 3 + y}{2} = \frac{7}{2}\]

\[ \Rightarrow y = 10\]  

\[ \frac{4 + z}{2} = - \frac{9}{2}\]

\[ \Rightarrow z = - 13\]

\[\text{ Position vector of point D } = \hat{i} + 10 \hat{j} - 13 \hat{k} \]The vector equation of line BD passing through the points with position vectors

\[\vec{a}\] (B) and \[\vec{b}\] (D) is 

\[\vec{r} = \vec{a} + \lambda \left( \vec{b} - \vec{a} \right)\]

Here,

\[\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \]

\[ \vec{b} = \hat{i} + 10 \hat{j} - 13 \hat{k} \]

Vector equation of the required line is

\[\vec{r} = \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left\{ \left( \hat{i} + 10 \hat{j} - 13 \hat{k} \right) - \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) \right\}\]

\[ \Rightarrow \vec{r} = \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda \left( - \hat{i} + 13 \hat{j} - 17 \hat{k}  \right) . . . (1) \]

\[\text{ Here }, \lambda \text{ is a parameter } . \]

Reducing (1) to cartesian form, we get

\[x \hat{i} + y \hat{ j} + z \hat{k} = \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda \left( - \hat{i} + 13 \hat{j} - 17 \hat{k} \right) [\text{ Putting } r^\to = x \hat{i} + y \hat{j} + z \hat{k}  \text
{ in } (1)]\]

\[ \Rightarrow x \hat {i} + y \hat{j} + z \hat{k} = \left( 2 - \lambda \right) \hat{i} + \left( - 3 + 13\lambda \right) \hat{j} + \left( 4 - 17\lambda \right) \hat{k}\]

\[\text{ Comparing the coefficients of }  \hat{i} , \hat{j}  \text{ and } \hat{k}  , \text{ we get } \]

\[x = 2 - \lambda, y = - 3 + 13\lambda, z = 4 - 17\lambda\]

\[ \Rightarrow \frac{x - 2}{- 1} = \lambda, \frac{y + 3}{13} = \lambda, \frac{z - 4}{- 17} = \lambda\]

\[ \Rightarrow \frac{x - 2}{- 1} = \frac{y + 3}{13} = \frac{z - 4}{- 17} = \lambda\]

\[ \Rightarrow \frac{x - 2}{1} = \frac{y + 3}{- 13} = \frac{z - 4}{17} = - \lambda\]

\[ \text{ Hence, the cartesian form of (1) is } \]

\[\frac{x - 2}{1} = \frac{y + 3}{- 13} = \frac{z - 4}{17}\]