If \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) are two non zero vectors inclined at an angle \(\theta\), then identify the correct option out of the given options (A) \(\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|}\) (B) \(\vec{a}\) and \(\vec{b}\) are perpendicular, if \(a_1b_1 + a_2b_2 + a_3b_3 = 0\) (C) \(\vec{a}\) and \(\vec{b}\) are perpendicular, \(\frac{a_1}{b_1} = \frac{a_2}{b_2} \neq \frac{c_1}{c_2}\) (D) for \(\theta = \pi\), \(\vec{a} \times \vec{b} = 0\) (E) \(\cos\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}|\cdot|\vec{b}|}\) Choose the most appropriate answer from the options given below

Distance between two planes \(x + 2y - z = 5\) and \(2x + 4y - 2z + 2 = 0\) is

If \(\vec{p} = \hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{q} = 2\hat{i} + \hat{j} - \hat{k}\), then the area of parallelogram having diagonals \((\vec{p} + \vec{q})\) and \((\vec{p} - \vec{q})\) is

The shortest distances of the point \((1,2,3)\) from \(x, y, z\) axes respectively are

If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three unit vectors such that \(\vec{a} + \vec{b} + \vec{c} = 0\) then the value of \(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\) is

\(\int\sqrt{1 - 49x^2}\,dx\) is equal to

The corner points of the feasible region for an L.P.P. are \((0, 10)\), \((5, 5)\), \((15, 15)\) and \((0, 20)\). If the objective function is \(z = px + qy\); \(p, q > 0\), then the condition on \(p\) and \(q\) so that the maximum of \(z\) occurs at \((15, 15)\) and \((0, 20)\)…

The order and degree of the differential equation \(\left[\left(\frac{d^2y}{dx^2}\right) - 3\right]^{\frac{1}{3}} = 2\left(\frac{dy}{dx}\right)^{\frac{1}{4}}\) are

\(\int x\sqrt{x + 2}\,dx\) is equal to: