Question 23 of 25intermediate🔧 ApplyShort Answer4 marks

If the lines x−1−3=y−22k=z−32\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2} and x−13k=y−11=z−6−5\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5} are perpendicular, find the value of kk.

Correct Answer

k = -10/7

Exercise: Miscellaneous Exercise on Chapter 11 | Q: 3 | (Page Page 15)
For More Understanding

Explanation

This question tests the concept of angle between two lines in 3D space. When two lines are perpendicular, the dot product of their direction vectors equals zero. The direction ratios are extracted from the symmetric form of line equations, and the perpendicularity condition is applied to find the unknown parameter k.

Solution Steps

  1. Step 1: Identify the direction vectors of both lines from the symmetric form. For Line 1: (x-1)/(-3) = (y-2)/(2k) = (z-3)/2, the direction vector is b₁ = -3î + 2kĵ + 2k̂. For Line 2: (x-1)/(3k) = (y-1)/1 = (z-6)/(-5), the direction vector is b₂ = 3kî + 1ĵ - 5k̂.

  2. Step 2: Apply the perpendicularity condition. Since the lines are perpendicular, b₁ · b₂ = 0.

  3. Step 3: Calculate the dot product: (-3)(3k) + (2k)(1) + (2)(-5) = -9k + 2k - 10 = -7k - 10.

  4. Step 4: Set the dot product equal to zero and solve: -7k - 10 = 0, which gives k = -10/7.

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