If vec{a}=4 hat{i}+13 hat{j}-18 hat{k},vec{b} | vedclass

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(D) Given $\vec{a}=x \vec{b}+y \vec{c}$.Substituting the vectors,we get:$4 \hat{i}+13 \hat{j}-18 \hat{k} = x(\hat{i}-2 \hat{j}+3 \hat{k}) + y(2 \hat{i

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(D) Given $\vec{a}=x \vec{b}+y \vec{c}$.Substituting the vectors,we get:$4 \hat{i}+13 \hat{j}-18 \hat{k} = x(\hat{i}-2 \hat{j}+3 \hat{k}) + y(2 \hat{i}+3 \hat{j}-4 \hat{k})$$= (x+2y) \hat{i} + (-2x+3y) \hat{j} + (3x-4y) \hat{k}$.Comparing the coefficients of $\hat{i}, \hat{j}, 	ext{ and } \hat{k}$ on both sides,we get the system of equations:$1) x + 2y = 4$$2) -2x + 3y = 13$$3) 3x - 4y = -18$From equation $(1)$,$x = 4 - 2y$.Substituting this into equation $(2)$:$-2(4 - 2y) + 3y = 13$$-8 + 4y + 3y = 13$$7y = 21 \Rightarrow y = 3$.Now,substituting $y = 3$ into $x = 4 - 2y$:$x = 4 - 2(3) = 4 - 6 = -2$.Checking with equation $(3)$:$3(-2) - 4(3) = -6 - 12 = -18$ (Verified).Thus,$x = -2$ and $y = 3$.Therefore,$x+y = -2 + 3 = 1$.

If $\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$,and $\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ are three vectors such that $\vec{a}=x \vec{b}+y \vec{c}$,then $x+y=$

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