Question 32 of 68advanced🔧 ApplyNumerical2 marks

If a⃗, b⃗, c⃗ are unit vectors such that a⃗ + b⃗ + c⃗ = 0⃗, find the value of a⃗ ⋅ b⃗ + b⃗ ⋅ c⃗ + c⃗ ⋅ a⃗.

Correct Answer

-32\frac{3}{2}

Exercise: EXERCISE 10.3 | Q: 13 | (Page 25)
For More Understanding

Explanation

The problem requires evaluating the sum of dot products of three unit vectors whose sum is the zero vector. By squaring the magnitude of the sum of the vectors, we utilize the expansion formula provided in the context to relate the magnitudes to the dot products. Since the vectors are unit vectors, their squared magnitudes are 1, allowing us to solve for the required expression algebraically.

Solution Steps

  1. Step 1: Square both sides of the given equation a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0}.

    ∣a⃗+b⃗+c⃗∣2=∣0⃗∣2=0|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{0}|^2 = 0

  2. Step 2: Expand the left-hand side using the dot product expansion shown in the context:

    (a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗)=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})

  3. Step 3: Substitute the values of unit vector magnitudes (∣a⃗∣=1,∣b⃗∣=1,∣c⃗∣=1|\vec{a}|=1, |\vec{b}|=1, |\vec{c}|=1) into the equation:

    1+1+1+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=01 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0

  4. Step 4: Solve for the required value:

    3+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=03 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=−32(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3 a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}
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