The scalar product of the vector vec{a} = hat | vedclass

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(B) Given vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$,and $\vec{c} = \lambda \hat{i} + 2 \hat{j}

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(B) Given vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$,and $\vec{c} = \lambda \hat{i} + 2 \hat{j} + 3 \hat{k}$.Let $\vec{v} = \vec{b} + \vec{c} = (\lambda + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}$.The unit vector along $\vec{v}$ is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{(\lambda + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{(\lambda + 2)^2 + 6^2 + (-2)^2}} = \frac{(\lambda + 2) \hat{i} + 6 \hat{j} - 2 \hat{k}}{\sqrt{(\lambda + 2)^2 + 40}}$.The scalar product of $\vec{a}$ and $\hat{u}$ is $1$,so $\vec{a} \cdot \hat{u} = 1$.$\frac{(\lambda + 2)(1) + (6)(1) + (-2)(1)}{\sqrt{(\lambda + 2)^2 + 40}} = 1$.$\frac{\lambda + 2 + 6 - 2}{\sqrt{(\lambda + 2)^2 + 40}} = 1$.$\frac{\lambda + 6}{\sqrt{(\lambda + 2)^2 + 40}} = 1$.$\lambda + 6 = \sqrt{(\lambda + 2)^2 + 40}$.Squaring both sides: $(\lambda + 6)^2 = (\lambda + 2)^2 + 40$.$\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 4 + 40$.$12\lambda + 36 = 4\lambda + 44$.$8\lambda = 8$.$\lambda = 1$.

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

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