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:

Cross product is the go-to tool when a question mentions being "perpendicular to both" two existing directions. If the question asked for Cartesian form, the origin point \((0,0,0)\) makes the denominators simple: \(x/6 = y/-1 = z/-7\).

Answer 1

Concept:
• A line perpendicular to two given lines has a direction parallel to the cross product of their direction vectors.
• Vector equation of a line passing through point \(\vec{a}\) with direction \(\vec{n}\) is \(\vec{r} = \vec{a} + k\vec{n}\).

Step 1: Extract direction vectors from given lines
Direction of line 1: \(\vec{b_1} = 3\hat{i} + 4\hat{j} + 2\hat{k}\). Direction of line 2: \(\vec{b_2} = \hat{i} - \hat{j} + \hat{k}\).

Step 2: Find the direction of the required line
The direction \(\vec{n}\) is the cross product of \(\vec{b_1}\) and \(\vec{b_2}\): \[ \vec{n} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 1 & -1 & 1 \end{vmatrix} \] \[ \vec{n} = \hat{i}(4 - (-2)) - \hat{j}(3 - 2) + \hat{k}(-3 - 4) \] \[ \vec{n} = 6\hat{i} - \hat{j} - 7\hat{k} \]

Step 3: Formulate the line equation
The line passes through the origin, so \(\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k}\). The vector equation is: \[ \vec{r} = \vec{0} + \lambda'(6\hat{i} - \hat{j} - 7\hat{k}) \] \[ \vec{r} = \lambda'(6\hat{i} - \hat{j} - 7\hat{k}) \]