Question

If the equation of the line passing through the point $ \left( 0, -\frac{1}{2}, 0 \right) $ and perpendicular to the lines $ \mathbf{r_1} = \lambda ( \hat{i} + a \hat{j} + b \hat{k}) \quad \text{and} \quad \mathbf{r_2} = ( \hat{i} - \hat{j} - 6 \hat{k} ) + \mu( -b \hat{i} + a \hat{j} + 5 \hat{k}), $ is $ \frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}, $ then $ a + b + c + d $ is equal to:

• A. 10
• B. 12
• C. 13
• D. 14

Hint:

When solving for lines and vectors, remember that perpendicular lines' direction ratios must satisfy certain conditions. Use the cross product to find the direction ratios of the required line.

Answer 1

The Correct Option is D

The line is perpendicular to the two given lines, so the required line will be parallel to the cross product of the direction ratios of the two lines. 
The direction ratios of the first line \( \mathbf{r_1} \) are \( (1, a, b) \), and the direction ratios of the second line \( \mathbf{r_2} \) are \( (-b, a, 5) \).
The cross product of these direction ratios gives the direction ratios of the required line. 
The cross product of \( (1, a, b) \) and \( (-b, a, 5) \) is: \[ \hat{i}(a \cdot 5 - b \cdot a) - \hat{j}(1 \cdot 5 - b \cdot 1) + \hat{k}(1 \cdot a - a \cdot (-b)) = \hat{i}(5a - ab) - \hat{j}(5 - b) + \hat{k}(a + ab). \] Thus, the direction ratios of the required line are \( (5a - ab, -(5 - b), a + ab) \). Let the direction ratios of the required line be \( (5a - ab, -(5 - b), a + ab) = \alpha(5a - ab, -(b^2 + 5), a + ab) \). 
Now, since the line passes through the point \( \left( 0, -\frac{1}{2}, 0 \right) \), we can use the parametric equations: \[ \frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}. \] Substituting the values into the equations, we find \( d = 7 \) and \( c = 2 \). 
Using the system of equations to find \( a \) and \( b \): From \( 5a - ab = \frac{b^2 + 5}{-2} \), we calculate \( b = 3 \), and \( a = 2 \). 
Finally, we calculate \( a + b + c + d = 2 + 3 + 2 + 7 = 14 \). 
Thus, the correct answer is \( 14 \), which corresponds to option (4).