Suppose (l_1, m_1, n_1) and (l_2, m_2, n_2) a | vedclass

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(D) Let the direction cosines ($DC$'s) of $AB$ and $AC$ be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively.First,find the direction ratios ($DR$'

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$\frac{1}{35}$

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(D) Let the direction cosines ($DC$'s) of $AB$ and $AC$ be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively.First,find the direction ratios ($DR$'s) of $AB$ and $AC$:$DR's$ of $AB = (3-1, 1-(-2), -3-3) = (2, 3, -6)$.$DR's$ of $AC = (-3-1, 1-(-2), 3-3) = (-4, 3, 0)$.Now,calculate the direction cosines by dividing the $DR$'s by their magnitudes:Magnitude of $AB = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.So,$l_1 = \frac{2}{7}, m_1 = \frac{3}{7}, n_1 = \frac{-6}{7}$.Magnitude of $AC = \sqrt{(-4)^2 + 3^2 + 0^2} = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$.So,$l_2 = \frac{-4}{5}, m_2 = \frac{3}{5}, n_2 = 0$.Since $\cos A = l_1 l_2 + m_1 m_2 + n_1 n_2$:$\cos A = \left(\frac{2}{7}\right)\left(\frac{-4}{5}\right) + \left(\frac{3}{7}\right)\left(\frac{3}{5}\right) + \left(\frac{-6}{7}\right)(0)$$\cos A = \frac{-8}{35} + \frac{9}{35} + 0 = \frac{1}{35}$.

Suppose $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines and $\theta$ is the angle between them,where $\cos \theta = \pm(l_1 l_2 + m_1 m_2 + n_1 n_2)$. Let $A=(1, -2, 3)$,$B=(3, 1, -3)$,and $C=(-3, 1, 3)$ be the vertices of $\triangle ABC$. Then,$\cos A =$

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