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: State the angle bisector condition.
Let c⃗ = a⃗ + b⃗. Since c⃗ bisects the angle between a⃗ and b⃗, the cosines of the angles it makes with a⃗ and b⃗ are equal: (a⃗·c⃗)/(|a⃗||c⃗|) = (b⃗·c⃗)/(|b⃗||c⃗|). Cancel |c⃗|.

Step 2: Substitute c⃗ = a⃗ + b⃗ and expand.
(a⃗·(a⃗ + b⃗))/|a⃗| = (b⃗·(a⃗ + b⃗))/|b⃗| → (|a⃗|² + a⃗·b⃗)/|a⃗| = (a⃗·b⃗ + |b⃗|²)/|b⃗|.

Step 3: Cross-multiply and rearrange.
|b⃗|(|a⃗|² + a⃗·b⃗) = |a⃗|(a⃗·b⃗ + |b⃗|²). Rearranging yields (|b⃗| - |a⃗|)[a⃗·b⃗ - |a⃗||b⃗|] = 0.

Step 4: Use the non-collinearity condition.
Since a⃗ and b⃗ are not collinear, their dot product is strictly less than the product of their magnitudes: a⃗·b⃗ ≠ |a⃗||b⃗|. Hence the second factor is non-zero.

Step 5: Conclude the necessary equality.
The first factor must vanish: |b⃗| - |a⃗| = 0, so |a⃗| = |b⃗|.

Step 6: Final conclusion.
The vectors have equal magnitudes.