# What is the angle between the two lines having direction ratios (6, 3, 6) and (3, 3, 0)?

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What is the angle between the two lines having direction ratios (6, 3, 6) and (3, 3, 0)?
1. $\dfrac{\pi}{6}$
2. $\dfrac{\pi}{4}$
3. $\dfrac{\pi}{3}$
4. $\dfrac{\pi}{2}$

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Correct Answer - Option 2 : $\dfrac{\pi}{4}$

CONCEPT:

The angle θ between two lines whose direction ratios are proportional to a1, b1, c1 and a2, b2, c2 respectively is given by: $\cos θ = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \;\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|$

CALCULATION:

Here, we have to find the angle between the two lines having direction ratios (6, 3, 6) and (3, 3, 0).

Here, a1 = 6, b1 = 3, c1 = 6, a2 = 3, b2 = 3 and c2 = 0

As we know that, if θ is the angle between two lines having direction ratios proportional to a1, b1, c1 and a2, b2, c2 is given by: $\cos θ = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \;\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|$

⇒ $\cos θ = \left| {\frac{{6 \cdot 3 + 3 \cdot 3 +6 \cdot 0}}{{\sqrt {6^2 + 3^2 + 6^2} \;\sqrt {3^2 + 3^2 + 0^2} }}} \right|$

⇒ $cos θ = \frac{1}{\sqrt 2}$

⇒ θ = π/4

Hence, correct option is 2.