Question

Let the position vectors of points A, B and C be \( \mathbf{a} = 3\hat{i} - \hat{j} - 2\hat{k} \), \( \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k} \), and \( \mathbf{c} = \hat{i} + 5\hat{j} + 3\hat{k} \), respectively. Find the vector and Cartesian equations of the line passing through \( A \) and parallel to line \( BC \).

Hint:

When finding the equation of a line passing through a point and parallel to a given line, first find the direction vector of the given line, then use the parametric equations to describe the line. Eliminate the parameter to obtain the Cartesian equation.

Answer 1

To determine the equation of the line passing through point \( A \) and parallel to line \( BC \), the direction vector of line \( BC \) is first calculated. The direction vector of \( BC \) is given by: \[ \mathbf{BC} = \mathbf{c} - \mathbf{b} = (\hat{i} + 5\hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = 3\hat{j} + 4\hat{k}. \] Consequently, the direction vector for the line through \( A \) parallel to \( BC \) is \( \mathbf{d} = 3\hat{j} + 4\hat{k} \). The parametric equations for the line passing through \( A(3, -1, -2) \) and parallel to \( BC \) are derived as: \[ x = 3 + 0t, \quad y = -1 + 3t, \quad z = -2 + 4t. \] The vector form of the parametric equation for this line is: \[ \mathbf{r} = (3 + 0t)\hat{i} + (-1 + 3t)\hat{j} + (-2 + 4t)\hat{k}. \] The Cartesian equation is obtained by eliminating the parameter \( t \) from the parametric equations. Solving for \( t \) in terms of \( y \) and \( z \) from the equations \( y = -1 + 3t \) and \( z = -2 + 4t \) yields: \[ t = \frac{y + 1}{3}, \quad t = \frac{z + 2}{4}. \] Equating these expressions for \( t \) results in: \[ \frac{y + 1}{3} = \frac{z + 2}{4}. \] Cross-multiplication simplifies this to: \[ 4(y + 1) = 3(z + 2), \] which further simplifies to: \[ 4y + 4 = 3z + 6 \quad \Rightarrow \quad 4y - 3z = 2. \] Therefore, the Cartesian equation of the line is: \[ 4y - 3z = 2. \]