The angle between the lines 3x = 3y = -2z and 2x = -y = -3z is:

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  1. 90°
  2. 30°
  3. 45°
  4. 60°

Answer (Detailed Solution Below)

Option 1 : 90°
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Detailed Solution

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Given:

The lines 3x = 3y = -2z and 2x = -y = -3z

Concept:

Use formula for angle between two vectors a and b 

\(\rm \cos\theta=\frac{a\cdot b}{|a||b|}\)

Calculation:

The lines 3x = 3y = -2z and 2x = -y = -3z

The the standard form is 

\(\rm \frac{x}{1/3}=\frac{y}{1/3}=\frac{z}{-1/2}\) and \(\rm \frac{x}{1/2}=\frac{y}{-1}=\frac{z}{-1/3}\)

Then the direction ratios of given lines are

\(\rm (\frac{1}{3},\frac{1}{3},\frac{-1}{2})\ \ and \ (\frac{1}{2}, -1, \frac{-1}{3})\)

Now let angle between lines is θ then

\(\rm \cosθ=\frac{(\frac{1}{3},\frac{1}{3},\frac{-1}{2})\ \cdot \ (\frac{1}{2}, -1, \frac{-1}{3})}{|(\frac{1}{3},\frac{1}{3},\frac{-1}{2})|\ \cdot \ |(\frac{1}{2}, -1, \frac{-1}{3})|}\)

\(\rm \cosθ=\frac{\frac{1}{6}-\frac{1}{3}+\frac{1}{6}}{\sqrt{\frac{1}{9}+\frac{1}{9}+\frac{1}{4}}\sqrt{\frac{1}{4}+1+\frac{1}{9}}}\)

cos θ = 0

⇒ θ = 90o

Hence the option (1) is correct.

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