Question

Find the vector equation of a line passing through the origin and perpendicular to both the lines \( \vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k}) \) and \( \vec{r} = \mu(\hat{i} - \hat{j} + \hat{k}) \).

Hint:

Whenever a line is specified as being perpendicular to two other lines simultaneously, its direction vector can always be found immediately by computing the cross product of their individual direction vectors.

Answer 1

Concept: The vector equation of a straight line passing through a point with position vector \( \vec{a} \) and parallel to a direction vector \( \vec{b} \) is given by: \[ \vec{r} = \vec{a} + \kappa \vec{b} \] Here, the line passes through the origin, so \( \vec{a} = \vec{0} \). The required line is perpendicular to two given lines, meaning its direction vector \( \vec{b} \) must be perpendicular to the direction vectors of both given lines. We can find this direction vector by calculating the cross product of the direction vectors of the two lines.

Step 1: Extract the direction vectors of the given lines.
From the line equations, their respective parallel direction vectors are: \[ \vec{b}_1 = 3\hat{i} + 4\hat{j} + 2\hat{k} \] \[ \vec{b}_2 = \hat{i} - \hat{j} + \hat{k} \]

Step 2: Compute the cross product \( \vec{b} = \vec{b}_1 \times \vec{b}_2 \).
Using the determinant method for cross products: \[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 3 & 4 & 2 1 & -1 & 1 \end{vmatrix} \] Expanding along the first row: \[ \vec{b} = \hat{i}((4)(1) - (2)(-1)) - \hat{j}((3)(1) - (2)(1)) + \hat{k}((3)(-1) - (4)(1)) \] \[ \vec{b} = \hat{i}(4 + 2) - \hat{j}(3 - 2) + \hat{k}(-3 - 4) \] \[ \vec{b} = 6\hat{i} - \hat{j} - 7\hat{k} \]

Step 3: Write down the final vector line equation.
Since the line passes through the origin \( \vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} \), the vector equation simplifies to: \[ \vec{r} = \kappa(6\hat{i} - \hat{j} - 7\hat{k}) \] where \( \kappa \) is a scalar parameter.