Let vec{a}=2 hat{i}+3 hat{j}+hat{k},vec{b}=4 | vedclass

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(D) Given vectors are $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$ and $\vec{r}=x \hat{i}+y

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$(1, 3, -2)$

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(D) Given vectors are $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$ and $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$.From the dot products,we get the following system of linear equations:$1) \vec{r} \cdot \vec{a} = 2x + 3y + z = 9$$2) \vec{r} \cdot \vec{b} = 4x + y = 7$$3) \vec{r} \cdot \vec{c} = x - 3y - 7z = 6$From $(2)$,$y = 7 - 4x$.Substitute $y$ into $(1)$ and $(3)$:$(1) \Rightarrow 2x + 3(7 - 4x) + z = 9 \Rightarrow 2x + 21 - 12x + z = 9 \Rightarrow -10x + z = -12 \Rightarrow z = 10x - 12$$(3) \Rightarrow x - 3(7 - 4x) - 7(10x - 12) = 6$$x - 21 + 12x - 70x + 84 = 6$$-57x + 63 = 6$$-57x = -57 \Rightarrow x = 1$Now,find $y$ and $z$:$y = 7 - 4(1) = 3$$z = 10(1) - 12 = -2$Thus,$(x, y, z) = (1, 3, -2)$.

Let $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$,$\vec{r} \cdot \vec{a}=9$,$\vec{r} \cdot \vec{b}=7$,$\vec{r} \cdot \vec{c}=6$,then $(x, y, z) = $

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