Question

Consider the lines $L_1 : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}$ and $L_2 : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$, then the unit vector perpendicular to both $L_1$ and $L_2$ is

• A. $\frac{-\hat{i}+7\hat{j}+5\hat{k}}{5\sqrt{3}}$
• B. $\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$
• C. $\frac{+\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$
• D. $\frac{\hat{i}+7\hat{j}+5\hat{k}}{5\sqrt{3}}$

Hint:

Vector Tip: The cross product of two direction vectors is the most direct way to find the normal vector that is simultaneously orthogonal to both lines. Always remember the alternating signs ($+, -, +$) when expanding a $3\times3$ determinant.

Answer 1

The Correct Option is B

Concept: 3D Geometry - Cross Product and Unit Vectors.

Step 1: Identify the direction vectors of the two lines. The direction vector of a line in the form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ is given by $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. For line $L_1$, the direction vector is $\vec{b_1} = 3\hat{i} + \hat{j} + 2\hat{k}$. For line $L_2$, the direction vector is $\vec{b_2} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Step 2: Calculate the cross product to find a perpendicular vector. A vector perpendicular to both $L_1$ and $L_2$ is obtained by taking the cross product of their direction vectors ($\vec{b_1} \times \vec{b_2}$). Set up the determinant: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 1 & 2
1 & 2 & 3 \end{vmatrix}$.

Step 3: Expand the determinant. Expand along the first row: $= \hat{i}((1)(3) - (2)(2)) - \hat{j}((3)(3) - (2)(1)) + \hat{k}((3)(2) - (1)(1))$. $= \hat{i}(3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 1)$. $= -1\hat{i} - 7\hat{j} + 5\hat{k}$. This vector is $-\hat{i} - 7\hat{j} + 5\hat{k}$.

Step 4: Calculate the magnitude of the resulting vector. The magnitude of the cross product vector is $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + (-7)^2 + (5)^2}$. $|\vec{b_1} \times \vec{b_2}| = \sqrt{1 + 49 + 25} = \sqrt{75}$. Simplify the radical: $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$.

Step 5: Form the final unit vector. The unit vector $\hat{n}$ is found by dividing the vector by its magnitude: $\hat{n} = \frac{\vec{b_1} \times \vec{b_2}}{|\vec{b_1} \times \vec{b_2}|}$. Substituting our results: $\hat{n} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$. $$ \therefore \text{The required unit vector is } \frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}. $$