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: \(\frac{2}{3}\sqrt{34}\)
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 