Question

The vector equation of the line passing through the point \( (1,2,-4) \) and perpendicular to the two lines \[ \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \] and \[ \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \] is _____

• A. \( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} - 3\hat{j} + 6\hat{k}) \)
• B. \( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \)
• C. \( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} - 6\hat{k}) \)
• D. \( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} - 3\hat{j} - 6\hat{k}) \)

Hint:

For a line perpendicular to two lines, always take cross product of their direction vectors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A line perpendicular to two given lines has a direction vector equal to the cross product of the direction vectors of the two lines: $\vec{b} = \vec{b_1} \times \vec{b_2}$.
Step 2: Formula Application:
$\vec{b_1} = (3, -16, 7)$ and $\vec{b_2} = (3, 8, -5)$. $\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -16 & 7
3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$ $\vec{b} = 24\hat{i} + 36\hat{j} + 72\hat{k}$.
Step 3: Explanation:
Simplify the direction vector by dividing by 12: $\vec{b'} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. The line passes through $(1, 2, -4)$, so its position vector is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$. Equation: $\vec{r} = \vec{a} + \lambda\vec{b'}$.
Step 4: Final Answer:
The equation is $\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$.