Question

If \(\vec{a},\vec{b}\) be two non-collinear vectors and the vector \(\vec{a}+\vec{b}\) bisects the angle between \(\vec{a}\) and \(\vec{b}\), then

• A. \[ |\vec{a}|=|\vec{b}| \]
• B. Angle between \(\vec{a}\) and \(\vec{b}\) is \(0^\circ\) or \(\pi\)
• C. \(\vec{a}\) and \(\vec{b}\) always form adjacent sides of a square
• D. \(\vec{a}\) and \(\vec{b}\) always form adjacent sides of a rectangle

Hint:

The vector \(\vec{a}+\vec{b}\) bisects the angle between \(\vec{a}\) and \(\vec{b}\) if and only if the magnitudes of the vectors are equal, i.e., \( |\vec{a}|=|\vec{b}| \).

Answer 1

The Correct Option is A

Step 1: Use the condition that \(\vec{a}+\vec{b}\) bisects the angle between \(\vec{a}\) and \(\vec{b}\).
Let \[ \vec{c}=\vec{a}+\vec{b}. \] Since \(\vec{c}\) bisects the angle between \(\vec{a}\) and \(\vec{b}\), \[ \angle(\vec{a},\vec{c}) = \angle(\vec{c},\vec{b}). \] Therefore, \[ \frac{\vec{a}\cdot\vec{c}}{|\vec{a}|\,|\vec{c}|} = \frac{\vec{b}\cdot\vec{c}}{|\vec{b}|\,|\vec{c}|}. \] Cancelling \(|\vec{c}|\), \[ \frac{\vec{a}\cdot\vec{c}}{|\vec{a}|} = \frac{\vec{b}\cdot\vec{c}}{|\vec{b}|}. \]

Step 2: Substitute \(\vec{c}=\vec{a}+\vec{b}\).
We have \[ \frac{\vec{a}\cdot(\vec{a}+\vec{b})}{|\vec{a}|} = \frac{\vec{b}\cdot(\vec{a}+\vec{b})}{|\vec{b}|}. \] Expanding, \[ \frac{|\vec{a}|^2+\vec{a}\cdot\vec{b}}{|\vec{a}|} = \frac{\vec{a}\cdot\vec{b}+|\vec{b}|^2}{|\vec{b}|}. \]

Step 3: Simplify the equation.
Multiplying throughout by \(|\vec{a}|\,|\vec{b}|\), \[ |\vec{b}|\left(|\vec{a}|^2+\vec{a}\cdot\vec{b}\right) = |\vec{a}|\left(\vec{a}\cdot\vec{b}+|\vec{b}|^2\right). \] Rearranging, \[ |\vec{a}|^2|\vec{b}|-|\vec{a}||\vec{b}|^2 + (\vec{a}\cdot\vec{b})(|\vec{b}|-|\vec{a}|)=0. \] Taking \((|\vec{b}|-|\vec{a}|)\) common, \[ (|\vec{b}|-|\vec{a}|) \left[ \vec{a}\cdot\vec{b} - |\vec{a}|\,|\vec{b}| \right] =0. \]

Step 4: Use the non-collinearity condition.
Since \(\vec{a}\) and \(\vec{b}\) are non-collinear, \[ \vec{a}\cdot\vec{b} \neq |\vec{a}|\,|\vec{b}|. \] Hence, \[ |\vec{b}|-|\vec{a}|=0. \] Therefore, \[ |\vec{a}|=|\vec{b}|. \]

Step 5: Final conclusion.
Thus, \[ \boxed{|\vec{a}|=|\vec{b}|} \] is the correct statement.