The value of b such that the scalar product o | vedclass

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(D) Let $\vec{a} = i + j + k$,$\vec{u} = 2i + 4j - 5k$,and $\vec{v} = bi + 2j + 3k$.The sum of the vectors is $\vec{s} = \vec{u} + \vec{v} = (2+b)i +

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$1$

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(D) Let $\vec{a} = i + j + k$,$\vec{u} = 2i + 4j - 5k$,and $\vec{v} = bi + 2j + 3k$.The sum of the vectors is $\vec{s} = \vec{u} + \vec{v} = (2+b)i + 6j - 2k$.The unit vector parallel to $\vec{s}$ is $\hat{s} = \frac{(2+b)i + 6j - 2k}{\sqrt{(2+b)^2 + 6^2 + (-2)^2}} = \frac{(2+b)i + 6j - 2k}{\sqrt{b^2 + 4b + 4 + 36 + 4}} = \frac{(2+b)i + 6j - 2k}{\sqrt{b^2 + 4b + 44}}$.The scalar product of $\vec{a}$ and $\hat{s}$ is $1$:$\vec{a} \cdot \hat{s} = 1 \Rightarrow \frac{(1)(2+b) + (1)(6) + (1)(-2)}{\sqrt{b^2 + 4b + 44}} = 1$.Simplifying the numerator: $2 + b + 6 - 2 = b + 6$.So,$\frac{b+6}{\sqrt{b^2 + 4b + 44}} = 1$.Squaring both sides: $(b+6)^2 = b^2 + 4b + 44$.$b^2 + 12b + 36 = b^2 + 4b + 44$.$8b = 8 \Rightarrow b = 1$.

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

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