Question

The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is

• A. $\frac{x+1}{5} = \frac{y-2}{-4} = \frac{z+3}{1}$
• B. $\frac{x+1}{5} = \frac{y+2}{4} = \frac{z+3}{1}$
• C. $\frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{-1}$
• D. $\frac{x+1}{1} = \frac{y-2}{4} = \frac{z-3}{3}$

Hint:

To find a vector perpendicular to two others, use the Cross Product.

Answer 1

The Correct Option is C

The equation of a line passing through a given point and perpendicular to two other lines can be derived using vector operations. Let's find the direction ratios of the required line.

1. First, identify the direction ratios of the given lines:
   • Line L₁: \(2i -3j -2k\)
   • Line L₂: \(-i + 2j + 3k\)
2. The direction vector of a line perpendicular to both L₁ and L₂ is given by the cross-product of their direction vectors.
3. Calculate the cross product \(\mathbf{d} = (2i - 3j - 2k) \times (-i + 2j + 3k)\):
   • Using the cross product formula: 
     \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -2 \\ -1 & 2 & 3 \end{vmatrix}\)
   • Compute the determinant: \(\mathbf{d} = \hat{i}((-3)(3) - (-2)(2)) - \hat{j}(2(3) - (-2)(-1)) + \hat{k}(2 \times 2 - (-3)(-1))\)
   • \(\mathbf{d} = \hat{i}(-9 + 4) - \hat{j}(6 - 2) + \hat{k}(4 - 3)\)
   • \(\mathbf{d} = -5i -4j + 1k\)
4. Thus, the direction ratios of the line required are \((5, 4, -1)\).
5. The line passes through the point \((-1, 2, 3)\). Using these, the equation of the line is given by:
   \(\frac{x + 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{-1}\)

Thus, the correct answer is \(\frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{-1}\).

After checking against the options provided, this solution corresponds to the third option:

$\frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{-1}$