Question

Find the equation of a line in vector and Cartesian form which passes through the point \( (1, 2, -4) \) and is perpendicular to the lines \[ \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}. \] and \[ \vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu (3\hat{i} + 8\hat{j} - 5\hat{k}). \]

Hint:

To find the equation of a line perpendicular to two other lines, compute the cross product of their direction vectors to get the direction of the required line.

Answer 1

We are given the lines and need to find a line that is perpendicular to both. The direction vectors of the given lines are: \[ \vec{d_1} = \langle 3, -16, 7 \rangle \quad \text{and} \quad \vec{d_2} = \langle 3, 8, -5 \rangle. \] The direction vector of the required line is the cross product of \( \vec{d_1} \) and \( \vec{d_2} \): \[ \vec{d} = \vec{d_1} \times \vec{d_2}. \] We compute the cross product: \[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i} \begin{vmatrix} -16 & 7
8 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 7
3 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -16
3 & 8 \end{vmatrix}. \] After calculating the determinants, we get: \[ \vec{d} = \langle -51, -6, 69 \rangle. \] The equation of the line in vector form is: \[ \vec{r} = \vec{r_0} + \lambda \vec{d}, \] where \( \vec{r_0} = \langle 1, 2, -4 \rangle \) and \( \vec{d} = \langle -51, -6, 69 \rangle \). Hence, the equation of the line is: \[ \vec{r} = \langle 1, 2, -4 \rangle + \lambda \langle -51, -6, 69 \rangle. \] In Cartesian form, the equation is: \[ \frac{x - 1}{-51} = \frac{y - 2}{-6} = \frac{z + 4}{69}. \]