The scalar product of the vector hat{i}+hat{j | vedclass

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(A) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.Let $\vec{b} = (2 \hat{i}+4 \hat{j}-5 \hat{k}) + (\lambda \hat{i}+2 \hat{j}+3 \hat{k}) = (2+\lambda) \hat{

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(A) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.Let $\vec{b} = (2 \hat{i}+4 \hat{j}-5 \hat{k}) + (\lambda \hat{i}+2 \hat{j}+3 \hat{k}) = (2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$.The unit vector along $\vec{b}$ is $\hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^{2}+6^{2}+(-2)^{2}}} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^{2}+4 \lambda+44}}$.The scalar product $\vec{a} \cdot \hat{b} = 1$.$\Rightarrow \frac{(2+\lambda)(1) + (6)(1) + (-2)(1)}{\sqrt{\lambda^{2}+4 \lambda+44}} = 1$.$\Rightarrow \frac{\lambda+6}{\sqrt{\lambda^{2}+4 \lambda+44}} = 1$.$\Rightarrow \lambda+6 = \sqrt{\lambda^{2}+4 \lambda+44}$.Squaring both sides: $(\lambda+6)^{2} = \lambda^{2}+4 \lambda+44$.$\Rightarrow \lambda^{2}+12 \lambda+36 = \lambda^{2}+4 \lambda+44$.$\Rightarrow 8 \lambda = 8$.$\Rightarrow \lambda = 1$.

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

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