Question

The lines $ \frac{x-1}{2} = \frac{y-4}{4} = \frac{z-2}{3} \quad \text{and} \quad \frac{1-x}{1} = \frac{y-2}{5} = \frac{3-z}{a} \quad \text{are perpendicular to each other, then} \ a \ \text{equals to} $

• A. -6
• B. 6
• C. \( \frac{22}{3} \)
• D. \( -\frac{22}{3} \)

Hint:

To find the value of \( a \) for perpendicular lines, equate the dot product of their direction ratios to zero.

Answer 1

The Correct Option is B

To find the value of \( a \) such that the given lines are perpendicular, we need to determine the direction ratios of each line and use the condition for perpendicularity.

The direction ratios of the first line \(\frac{x-1}{2} = \frac{y-4}{4} = \frac{z-2}{3}\) are \(2, 4, 3\).

The direction ratios of the second line \(\frac{1-x}{1} = \frac{y-2}{5} = \frac{3-z}{a}\) can be rewritten for consistency in format: \(-1, 5, -a\).

The condition for two lines to be perpendicular in 3D space is that the dot product of their direction ratios is zero:

\(2(-1) + 4(5) + 3(-a) = 0\)

Simplifying, we get:

\(-2 + 20 - 3a = 0\)

Solving for \(a\):

\(18 = 3a\)

\(a = \frac{18}{3} = 6\)

Therefore, the correct value of \( a \) is 6.

Answer 2

The condition for two lines to be perpendicular is that the dot product of their direction ratios must be zero. 
For the first line, we have the direction ratios as \( \mathbf{l_1} = (2, 4, 3) \). For the second line, we have the direction ratios as \( \mathbf{l_2} = (-1, 5, -a) \). The condition for perpendicularity is: \[ 2 \cdot (-1) + 4 \cdot 5 + 3 \cdot (-a) = 0 \] Simplifying this: \[ -2 + 20 - 3a = 0 \] \[ 18 - 3a = 0 \] \[ 3a = 18 \] \[ a = 6 \] Thus, the value of \( a \) is 6.