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If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two vectors and θ is the angle between them, then the value of cos θ is

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$\large {\ell _1}{\ell _2} + {m_1}{m_2} + {n_1}{n_2}$

$\large {\ell _1}{m_1} + {m_1}{n_1} + {n_1}{\ell _1}$

$\large {\ell _2}{m_2} + {m_2}{n_2} + {n_2}{l_2}$

$\large {m_1}{\ell _2} + {\ell _2}{m_2} + {n_1}{m_2}$

answer is A.

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Detailed Solution

$\large l_1^2 + m_1^2 + n_1^2 = 1\;\&\; l_2^2 + m_2^2 + n_2^2 = 1$
$\large \vec A=A(l_1\hat i+m_1\hat j+n_1\hat k)\;and\; \vec B=B(l_2\hat i+m_2\hat j+n_2\hat k)$
Now, $\large cos\theta=\frac{\vec A.\vec B}{AB}=(l_1l_2+m_1m_2+n_1n_2)$

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