The scalar product of vectors bar{a}=hat{i}+2 | vedclass

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(C) Let $\bar{s} = \bar{b} + \bar{c} = (2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}$.The unit vector along $\bar{s}$ is $\hat{s} = \frac{\bar{s}}{|\bar

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$-3$

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(C) Let $\bar{s} = \bar{b} + \bar{c} = (2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}$.The unit vector along $\bar{s}$ is $\hat{s} = \frac{\bar{s}}{|\bar{s}|} = \frac{(2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{(2+\lambda)^2 + (-2)^2 + 2^2}} = \frac{(2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{\lambda^2 + 4\lambda + 12}}$.Given that $\bar{a} \cdot \hat{s} = 1$,we have:$(\hat{i} + 2 \hat{j} + \hat{k}) \cdot \left( \frac{(2+\lambda) \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{\lambda^2 + 4\lambda + 12}} \right) = 1$.Taking the dot product in the numerator:$\frac{(2+\lambda) - 4 + 2}{\sqrt{\lambda^2 + 4\lambda + 12}} = 1$.$\frac{\lambda}{\sqrt{\lambda^2 + 4\lambda + 12}} = 1$.Squaring both sides:$\lambda^2 = \lambda^2 + 4\lambda + 12$.$4\lambda = -12$.$\lambda = -3$.

The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\bar{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\bar{c}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ is $1$. Then the value of $\lambda$ is:

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