The angle between the lines 3x + 2y + z = 0 = | vedclass

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(B) The direction ratios $(l_1, m_1, n_1)$ of the first line are given by the cross product of the normals to the planes $3x + 2y + z = 0$ and $x + y

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$\frac{\pi}{2}$

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(B) The direction ratios $(l_1, m_1, n_1)$ of the first line are given by the cross product of the normals to the planes $3x + 2y + z = 0$ and $x + y - 2z = 0$.Using the cross product formula:$l_1 = (2)(-2) - (1)(1) = -5$$m_1 = (1)(1) - (3)(-2) = 7$$n_1 = (3)(1) - (2)(1) = 1$So,the direction ratios are proportional to $(-5, 7, 1)$ or $(5, -7, -1)$.The direction ratios $(l_2, m_2, n_2)$ of the second line are given by the cross product of the normals to the planes $2x - y - z = 0$ and $7x + 10y - 8z = 0$.$l_2 = (-1)(-8) - (-1)(10) = 8 + 10 = 18$$m_2 = (-1)(7) - (2)(-8) = -7 + 16 = 9$$n_2 = (2)(10) - (-1)(7) = 20 + 7 = 27$Dividing by $9$,we get the direction ratios as $(2, 1, 3)$.Let $	heta$ be the angle between the two lines. The cosine of the angle is given by:$\cos 	heta = \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}}$$\cos 	heta = \frac{|(5)(2) + (-7)(1) + (-1)(3)|}{\sqrt{5^2 + (-7)^2 + (-1)^2} \sqrt{2^2 + 1^2 + 3^2}}$$\cos 	heta = \frac{|10 - 7 - 3|}{\sqrt{25 + 49 + 1} \sqrt{4 + 1 + 9}} = \frac{0}{\sqrt{75} \sqrt{14}} = 0$Since $\cos 	heta = 0$,the angle $	heta = \frac{\pi}{2}$.

The angle between the lines $3x + 2y + z = 0 = x + y - 2z$ and $2x - y - z = 0 = 7x + 10y - 8z$ is

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