Question 66 of 68intermediate🔧 ApplyMCQ1 mark

Let a⃗ and b⃗ be two unit vectors and θ is the angle between them. Then a⃗ + b⃗ is a unit vector if

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 10 | Q: 17 | (Page 36)
For More Understanding

Explanation

Option D is correct. We are given that a⃗\vec{a} and b⃗\vec{b} are unit vectors, so ∣a⃗∣=1|\vec{a}| = 1 and ∣b⃗∣=1|\vec{b}| = 1. For the vector a⃗+b⃗\vec{a} + \vec{b} to be a unit vector, its magnitude must be equal to 1. Using the definition of the dot product provided in the context, a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta, and the property a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2, we can solve for the angle θ\theta. The calculation yields cos⁡θ=−1/2\cos \theta = -1/2, which corresponds to the angle θ=2π/3\theta = 2\pi/3.

Solution Steps

  1. Step 1: Let ∣a⃗+b⃗∣=1|\vec{a} + \vec{b}| = 1 since a⃗+b⃗\vec{a} + \vec{b} is a unit vector. Squaring both sides gives ∣a⃗+b⃗∣2=1|\vec{a} + \vec{b}|^2 = 1.

  2. Step 2: Expand the magnitude squared using the dot product property: ∣a⃗+b⃗∣2=(a⃗+b⃗)⋅(a⃗+b⃗)=∣a⃗∣2+∣b⃗∣2+2(a⃗⋅b⃗)|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}).

  3. Step 3: Substitute the given values ∣a⃗∣=1|\vec{a}| = 1 and ∣b⃗∣=1|\vec{b}| = 1. The equation becomes 1=12+12+2(a⃗⋅b⃗)1 = 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b}).

  4. Step 4: Use the context definition a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=(1)(1)cos⁡θ=cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta = (1)(1)\cos \theta = \cos \theta. Substitute this into the equation: 1=1+1+2cos⁡θ1 = 1 + 1 + 2\cos \theta.

  5. Step 5: Simplify to find cos⁡θ\cos \theta: 1=2+2cos⁡θ  ⟹  −1=2cos⁡θ  ⟹  cos⁡θ=−1/21 = 2 + 2\cos \theta \implies -1 = 2\cos \theta \implies \cos \theta = -1/2.

  6. Step 6: The angle θ\theta for which cos⁡θ=−1/2\cos \theta = -1/2 is 2π/32\pi/3. Thus, the correct option is (D).

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