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If $\mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4)$ and $\mathbf{i}$ is the unit vector in the $x$-direction, then $(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) . \mathbf{i}=$
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The correct answer is:
11
$\begin{aligned}
& \mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4) \\
& \text { So, } \mathbf{a}=(1,-1,2)=\mathbf{i}-\mathbf{j}+2 \mathbf{k} ; \mathbf{b}=(-2,3,5)=-2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k} \\
& \text { and } \mathbf{c}=(2,-2,4)=2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k} \\
& \Rightarrow \mathbf{a}-2 \mathbf{b}+3 \mathbf{c}=(\mathbf{i}-\mathbf{j}+2 \mathbf{k})-2(-2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k})
\end{aligned}$
$+3(2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k})$
$=11 \mathbf{i}-13 \mathbf{j}+4 \mathbf{k}$ and $(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) \cdot \mathbf{i}=11$.
& \mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4) \\
& \text { So, } \mathbf{a}=(1,-1,2)=\mathbf{i}-\mathbf{j}+2 \mathbf{k} ; \mathbf{b}=(-2,3,5)=-2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k} \\
& \text { and } \mathbf{c}=(2,-2,4)=2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k} \\
& \Rightarrow \mathbf{a}-2 \mathbf{b}+3 \mathbf{c}=(\mathbf{i}-\mathbf{j}+2 \mathbf{k})-2(-2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k})
\end{aligned}$
$+3(2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k})$
$=11 \mathbf{i}-13 \mathbf{j}+4 \mathbf{k}$ and $(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) \cdot \mathbf{i}=11$.
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