Question 68 of 68intermediate🔧 ApplyMCQ1 mark

If θ is the angle between any two vectors a⃗ and b⃗, then |a⃗ ⋅ b⃗| = |a⃗ × b⃗| when θ is equal to

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 10 | Q: 19 | (Page 36)
For More Understanding

Explanation

Option B is correct. From the context, the scalar product is defined as a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, and the cross product magnitude is ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta. For the given condition ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|, we get ∣a⃗∣∣b⃗∣∣cos⁡θ∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}| |\vec{b}| |\cos \theta| = |\vec{a}| |\vec{b}| \sin \theta. Simplifying (for non-zero vectors), ∣cos⁡θ∣=sin⁡θ|\cos \theta| = \sin \theta. At θ=π4\theta = \frac{\pi}{4}, we have ∣cos⁡(π4)∣=12|\cos(\frac{\pi}{4})| = \frac{1}{\sqrt{2}} and sin⁡(π4)=12\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}, satisfying the equation. Options A (θ=0), C (θ=π/2), and D (θ=π) do not satisfy this condition.

Solution Steps

  1. Step 1: From the context, ∣a⃗⋅b⃗∣=∣a⃗∣∣b⃗∣∣cos⁡θ∣|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta| and ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta.

  2. Step 2: Equating both expressions: ∣a⃗∣∣b⃗∣∣cos⁡θ∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}| |\vec{b}| |\cos \theta| = |\vec{a}| |\vec{b}| \sin \theta.

  3. Step 3: Simplifying: ∣cos⁡θ∣=sin⁡θ|\cos \theta| = \sin \theta.

  4. Step 4: At θ=π4\theta = \frac{\pi}{4}: ∣cos⁡(π4)∣=12=sin⁡(π4)|\cos(\frac{\pi}{4})| = \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4}). Condition satisfied.

  5. Step 5: Checking other options: θ=0 gives 1≠\neq0, θ=π/2 gives 0≠\neq1, θ=π gives 1≠\neq0. Only option B works.

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