Question:medium

The value of \( \lambda \) for which the lines \(\frac{2 - x}{3} = \frac{3 - 4y}{5} = \frac{z - 2}{3}\) and  \(\frac{x - 2}{-3} = \frac{2y - 4}{3} = \frac{2 - z}{\lambda}\) are perpendicular is:

Updated On: Mar 27, 2026
  • \(-2\)
  • \(2\)
  • \( \frac{8}{19} \)
  • \( \frac{19}{8} \)
Show Solution

The Correct Option is A

Solution and Explanation

To confirm perpendicularity of lines, calculate the dot product of their direction vectors and equate it to zero.

Line 1's direction vector is \( \vec{d_1} = \langle -3, -4, 3 \rangle \).

Line 2's direction vector is \( \vec{d_2} = \langle -3, 3, -\lambda \rangle \).

The dot product \( \vec{d_1} \cdot \vec{d_2} \) is computed as:

\[ \vec{d_1} \cdot \vec{d_2} = (-3)(-3) + (-4)(3) + (3)(-\lambda) \]

This simplifies to:

\[ \vec{d_1} \cdot \vec{d_2} = 6 - 12 - 3\lambda \]

For perpendicular lines, the condition is:

\[ 6 - 12 - 3\lambda = 0 \]

Further simplification yields:

\[ -6 - 3\lambda = 0 \implies -3\lambda = 6 \implies \lambda = -2 \]

Thus, the value of \( \lambda \) for perpendicular lines is:

\[ \boxed{\lambda = -2} \]

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