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Let the vectors  \vec a,\vec b\;\text{ and }\vec c such that \left | \vec a \right |=2,\left | \vec b \right |=4,\left | \vec c \right |=4. If the projection of \vec b on \vec a is equal to the projection of \vec c on \vec a and \vec b is \perp\;\vec c then the value of  \left | \vec a+\vec b-\vec c \right | = ?
Option: 1 \sqrt6
Option: 2 6
Option: 3 3
Option: 4 \sqrt3

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$ Given $ |\vec{a}| = 2, |\vec{b}| = 4, |\vec{c}| = 5 \\ \vec{b} \cdot \hat{a} =\vec{c} \cdot \hat{a} \Rightarrow \vec{a} \cdot \vec{b} = \vec{a}\cdot \vec{c} \\ \vec{b} \cdot \vec{c} = 0 \\ |\vec{a} + \vec{b} - \vec{c}|^2 = (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{a} + \vec{b} - \vec{c}) \\ = |\vec{a}|^2 + \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{b} + |\vec{b}|^2 - \vec{b} \cdot \vec{c}-\vec{a} \cdot \vec{c}- \vec{b} \cdot \vec{c} + |\vec{c}|^2 \\ = |\vec{a}|^2 + |\vec{b}|^2 +|\vec{c}|^2 \\ |\vec{a} + \vec{b} - \vec{c}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 +|\vec{c}|^2} \\ = \sqrt{2^2 + 4^2 + 4^2} \\ = 6

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