The given problem involves determining the point where the internal angle bisector of \(\angle AOB\) meets line AB. Let's solve this geometrically using vectors.
Given:
To find the length OC, where C is the point of intersection of the angle bisector of \(\angle AOB\) with the line AB, follow these steps:
The point C dividing the segment AB in the ratio \(|\mathbf{b}|:|\mathbf{a}|\) will be on the angle bisector of \(\angle AOB\).
Therefore, the length of OC is \(\frac{2}{3}\sqrt{34}\). This matches with the given correct option: \(\frac{2}{3}\sqrt{34}\).
A = (2, 2, 1) and B = (2, 4, 4)
The internal bisector of ∠AOB divides AB in the ratio OA : OB = 1 : 2. Using the section formula, the coordinates of C are:
\[ C = \frac{1 \cdot B + 2 \cdot A}{1 + 2} = \frac{1 \cdot (2, 4, 4) + 2 \cdot (2, 2, 1)}{3} = \left(2, \frac{8}{3}, 2\right) \]
The vector OC has coordinates (2, \(\frac{8}{3}\), 2). Using the distance formula:
\[ |OC| = \sqrt{2^2 + \left(\frac{8}{3}\right)^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{\frac{136}{9}} = \frac{2\sqrt{34}}{3} \]
So, the correct answer is: \(\frac{2}{3}\sqrt{34}\)
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______. 