Question 15 of 25intermediate🔧 ApplyShort Answer4 marks

Find the values of pp so that the lines 1−x3=7y−142p=z−32\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2} and 7−7x3p=y−51=6−z5\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} are at right angles.

Correct Answer

p = 70/11

Exercise: EXERCISE 11.2 | Q: 10 | (Page Page 14)
For More Understanding

Explanation

This question tests the concept of finding the angle between two lines using direction ratios. When two lines are perpendicular (at right angles), the angle between them is 90°, and cos 90° = 0. From the formula provided in the context, for perpendicular lines, the dot product of their direction ratios must equal zero: a₁a₂ + b₁b₂ + c₁c₂ = 0. The key skill is converting the given equations to standard symmetric form to extract direction ratios correctly.

Solution Steps

  1. Step 1: Convert Line 1 to standard form. Line 1: (1-x)/3 = (7y-14)/(2p) = (z-3)/2. Rewriting: (x-1)/(-3) = (y-2)/(2p/7) = (z-3)/2. Direction ratios: a₁ = -3, b₁ = 2p/7, c₁ = 2

  2. Step 2: Convert Line 2 to standard form. Line 2: (7-7x)/(3p) = (y-5)/1 = (6-z)/5. Rewriting: (x-1)/(-3p/7) = (y-5)/1 = (z-6)/(-5). Direction ratios: a₂ = -3p/7, b₂ = 1, c₂ = -5

  3. Step 3: Apply condition for perpendicular lines. For lines at right angles: a₁a₂ + b₁b₂ + c₁c₂ = 0

  4. Step 4: Substitute direction ratios. (-3)(-3p/7) + (2p/7)(1) + (2)(-5) = 0

  5. Step 5: Simplify and solve. 9p/7 + 2p/7 - 10 = 0, which gives 11p/7 = 10, therefore p = 70/11

PreviousNext