Question

The equation of the line passing through the point $(1,2,3)$ and perpendicular to the lines $\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z-3}{3}$ and $\vec r = \lambda(-3\hat i + 2\hat j + 5\hat k)$ is

• A. $\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(2\hat i + 7\hat j - 4\hat k)$
• B. $\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(2\hat i + 7\hat j + 4\hat k)$
• C. $\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(2\hat i - 7\hat j - 4\hat k)$
• D. $\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(2\hat i - 7\hat j + 4\hat k)$

Hint:

A line perpendicular to two given lines has a direction vector equal to the cross product of their direction vectors.

Answer 1

The Correct Option is D

Step 1: Finding direction vectors of the given lines.
From the symmetric form, the direction vector of the first line is \[ \vec d_1 = \langle 1,2,3 \rangle \] From the vector equation, the direction vector of the second line is \[ \vec d_2 = \langle -3,2,5 \rangle \]
Step 2: Finding a direction vector perpendicular to both lines.
The required direction vector is given by the cross product: \[ \vec d = \vec d_1 \times \vec d_2 \] \[ = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 3\\ -3 & 2 & 5 \end{vmatrix} = \langle 2, -7, 4 \rangle \]
Step 3: Writing the equation of the required line.
The line passes through $(1,2,3)$ and has direction vector $\langle 2,-7,4 \rangle$.
Step 4: Conclusion.
The required equation of the line is option (D).