Find the angle between the lines \[ \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{-5} \quad \text{and} \quad \frac{x+3}{-3} = \frac{y-1}{2} = \frac{z-5}{5}. \]
Step 1: The direction ratios of the first line are \( (-1, 2, -5) \) and for the second line, they are \( (-3, 2, 5) \).
Step 2: The angle \( \theta \) between the two lines is given by the formula: \[ \cos \theta = \frac{l_1 l_2}{|l_1| |l_2|} \] where \( l_1 \) and \( l_2 \) are the direction ratios of the two lines.
Step 3: Substitute the values of \( l_1 = (-1, 2, -5) \) and \( l_2 = (-3, 2, 5) \) into the formula and calculate: \[ \cos \theta = \frac{(-1)(-3) + (2)(2) + (-5)(5)}{\sqrt{(-1)^2 + 2^2 + (-5)^2} \times \sqrt{(-3)^2 + 2^2 + 5^2}}. \]
Step 4: Simplify the equation and solve for \( \theta \).
Show that the vectors \( 2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k}, 3\hat{i} - 4\hat{j} - 4\hat{k} \) form the vertices of a right-angled triangle.
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $