If vec{a} = hat{i} + hat{j} + hat{k} and vec{ | vedclass

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(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$.The dot product is $\vec{a} \cdot \vec{b} = (1)(x) +

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$36$

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(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$.The dot product is $\vec{a} \cdot \vec{b} = (1)(x) + (1)(y) + (1)(z) = x + y + z$.We are given $\vec{a} \cdot \vec{b} = 10$,so $x + y + z = 10$.Since $x, y, z \in \mathbb{N}$ (natural numbers,where $x, y, z \ge 1$),we use the stars and bars formula for the number of positive integer solutions to the equation $x_1 + x_2 + \dots + x_k = n$,which is given by $\binom{n-1}{k-1}$.Here,$n = 10$ and $k = 3$.Number of solutions = $\binom{10-1}{3-1} = \binom{9}{2} = \frac{9 	imes 8}{2 	imes 1} = 36$.

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$,find the number of possible vectors $\vec{b}$ such that $\vec{a} \cdot \vec{b} = 10$,where $(x, y, z) \in \mathbb{N}$.

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