If vec{a} = 2hat{i} + 2hat{j} - hat{k}, vec{b | vedclass

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(D) Given $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.Using the property $

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(D) Given $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.Using the property $|\vec{u} \pm \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2(\vec{u} \cdot \vec{v})$, we have $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.This simplifies to $4(\vec{a} \cdot \vec{b}) = 0$, which implies $\vec{a} \cdot \vec{b} = 0$.Calculating the dot product: $(2\hat{i} + 2\hat{j} - \hat{k}) \cdot (\alpha\hat{i} + \beta\hat{j} + 2\hat{k}) = 0$.$2\alpha + 2\beta - 2 = 0$.Dividing by $2$, we get $\alpha + \beta - 1 = 0$, so $\alpha + \beta = 1$.

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}$ and $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, then $\alpha + \beta$ is equal to

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