Skip to main contentSkip to solution

If a⃗\vec{a} and b⃗\vec{b} are two non-zero vectors such that ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|, then the angle θ\theta between a⃗\vec{a} and b⃗\vec{b} is

Solution

✅ Correct Option: 2

Given two non-zero vectors a⃗\vec{a} and b⃗\vec{b} such that:

∣a⃗⋅b⃗∣=∣a⃗×b⃗∣|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|

The magnitude of the dot product equals the magnitude of the cross product.


The dot product formula is:

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta

The cross product formula is:

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta

where θ\theta is the angle between the vectors.


Substituting into the given condition:

∣∣a⃗∣∣b⃗∣cos⁡θ∣=∣a⃗∣∣b⃗∣sin⁡θ||\vec{a}||\vec{b}|\cos\theta| = |\vec{a}||\vec{b}|\sin\theta


Since a⃗\vec{a} and b⃗\vec{b} are non-zero vectors, ∣a⃗∣∣b⃗∣≠0|\vec{a}||\vec{b}| \neq 0.

Dividing both sides by ∣a⃗∣∣b⃗∣|\vec{a}||\vec{b}|:

∣cos⁡θ∣=sin⁡θ|\cos\theta| = \sin\theta


For angles between vectors, 0≤θ≤π0 \leq \theta \leq \pi, which means sin⁡θ≥0\sin\theta \geq 0.

For 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}, cos⁡θ\cos\theta is positive, so ∣cos⁡θ∣=cos⁡θ|\cos\theta| = \cos\theta.

This gives:

cos⁡θ=sin⁡θ\cos\theta = \sin\theta

1=sin⁡θcos⁡θ1 = \frac{\sin\theta}{\cos\theta}

1=tan⁡θ1 = \tan\theta

θ=π4\theta = \frac{\pi}{4}

Therefore, the angle between a⃗\vec{a} and b⃗\vec{b} is π4\frac{\pi}{4} or 45°45°.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question