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

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(D) Let $\vec{s} = \vec{u} + \vec{v} = (2+b)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k} = (2+b)\hat{i} + 6\hat{j} - 2\hat{k}$.The unit vector parallel to $

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$1$

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(D) Let $\vec{s} = \vec{u} + \vec{v} = (2+b)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k} = (2+b)\hat{i} + 6\hat{j} - 2\hat{k}$.The unit vector parallel to $\vec{s}$ is $\hat{s} = \frac{\vec{s}}{|\vec{s}|} = \frac{(2+b)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2+b)^2 + 6^2 + (-2)^2}}$.The scalar product of $\vec{a}$ and $\hat{s}$ is $1$, so $\vec{a} \cdot \hat{s} = 1$.$\frac{(1)(2+b) + (1)(6) + (1)(-2)}{\sqrt{(2+b)^2 + 36 + 4}} = 1$.$\frac{b+6}{\sqrt{(2+b)^2 + 40}} = 1$.Squaring both sides: $(b+6)^2 = (2+b)^2 + 40$.$b^2 + 12b + 36 = b^2 + 4b + 4 + 40$.$8b = 8$, so $b = 1$.

The value of $b$ such that the scalar product of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with the unit vector parallel to the sum of the vectors $\vec{u} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v} = b\hat{i} + 2\hat{j} + 3\hat{k}$ is $1$, is...

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