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The value of $b$ such that scalar product of the vectors $(\mathbf{i}+\mathbf{j}+\mathbf{k})$ with the unit vector parallel to the sum of the vectors $(2 \mathbf{i}+4 \mathbf{j}-\mathbf{5})$ and $(b \mathbf{i}+\mathbf{j}+3 \mathbf{k})$ is 1 , is
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The correct answer is:
1
$\begin{aligned}
& \text { Parallel vector }=(2+b) \mathbf{i}+6 \mathbf{j}-2 \mathbf{k} \\
& \text { Unit vector }=\frac{(2+b) \mathbf{i}+6 \mathbf{j}-2 \mathbf{k}}{\sqrt{b^2+4 b+44}}
\end{aligned}$
According to the condition,
$\begin{aligned}
& \qquad 1=\frac{(2+b)+6-2}{\sqrt{b^2+4 b+44}} \\
& \text { According to the condition, } \\
& \Rightarrow b^2+4 b+44=b^2+12 b+36 \Rightarrow 8 b=8 \Rightarrow b=1 .
\end{aligned}$
& \text { Parallel vector }=(2+b) \mathbf{i}+6 \mathbf{j}-2 \mathbf{k} \\
& \text { Unit vector }=\frac{(2+b) \mathbf{i}+6 \mathbf{j}-2 \mathbf{k}}{\sqrt{b^2+4 b+44}}
\end{aligned}$
According to the condition,
$\begin{aligned}
& \qquad 1=\frac{(2+b)+6-2}{\sqrt{b^2+4 b+44}} \\
& \text { According to the condition, } \\
& \Rightarrow b^2+4 b+44=b^2+12 b+36 \Rightarrow 8 b=8 \Rightarrow b=1 .
\end{aligned}$
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