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JEE Advanced 2012 · Vector Algebra

If a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} are unit vectors satisfying ∣a⃗−b⃗∣2+∣b⃗−c⃗∣2+∣c⃗−a⃗∣2=9|\vec{a}-\vec{b}|^{2} + |\vec{b}-\vec{c}|^{2} + |\vec{c}-\vec{a}|^{2} = 9, then ∣2a⃗+5b⃗+5c⃗∣|2\vec{a}+5\vec{b}+5\vec{c}| is

Answer: 33

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Written solution

∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec{a}| = |\vec{b}| = |\vec{c}| = 1

∣a⃗−b⃗∣2+∣b⃗−c⃗∣2+∣c⃗−a⃗∣2=9|\vec{a}-\vec{b}|^{2} + |\vec{b}-\vec{c}|^{2} + |\vec{c}-\vec{a}|^{2} = 9

∣2a⃗+5b⃗+5c⃗∣=?|2\vec{a} + 5\vec{b} + 5\vec{c}| = ?

1. expand ∣a⃗−b⃗∣2 with dot products1.\ \text{expand } |\vec{a}-\vec{b}|^2 \text{ with dot products}

2. find a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗2.\ \text{find } \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}

3. get a⃗+b⃗+c⃗, then the answer3.\ \text{get } \vec{a}+\vec{b}+\vec{c}, \text{ then the answer}

∣a⃗−b⃗∣2=2−2 a⃗⋅b⃗|\vec{a}-\vec{b}|^2 = 2 - 2\,\vec{a}\cdot\vec{b}

∣b⃗−c⃗∣2=2−2 b⃗⋅c⃗,∣c⃗−a⃗∣2=2−2 c⃗⋅a⃗|\vec{b}-\vec{c}|^2 = 2 - 2\,\vec{b}\cdot\vec{c}, \quad |\vec{c}-\vec{a}|^2 = 2 - 2\,\vec{c}\cdot\vec{a}

6−2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=96 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9

⇒a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−32\Rightarrow \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\dfrac{3}{2}

∣a⃗+b⃗+c⃗∣2=3+2(−32)=0|\vec{a}+\vec{b}+\vec{c}|^2 = 3 + 2\left(-\dfrac{3}{2}\right) = 0

⇒a⃗+b⃗+c⃗=0⃗\Rightarrow \vec{a}+\vec{b}+\vec{c} = \vec{0}

2a⃗+5b⃗+5c⃗=2a⃗+5(b⃗+c⃗)=2a⃗−5a⃗=−3a⃗2\vec{a}+5\vec{b}+5\vec{c} = 2\vec{a} + 5(\vec{b}+\vec{c}) = 2\vec{a} - 5\vec{a} = -3\vec{a}

∣2a⃗+5b⃗+5c⃗∣=∣−3a⃗∣=3∣a⃗∣=3|2\vec{a}+5\vec{b}+5\vec{c}| = |-3\vec{a}| = 3|\vec{a}| = 3

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