Let vec{a}=hat{i}+2 hat{j}+3 hat{k},vec{b}=2 | vedclass

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(B) Given $\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$ and $\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$.Then $\vec{b} + \vec{c} = (2+3)\hat{i} + (3-1

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$25$

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(B) Given $\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$ and $\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$.Then $\vec{b} + \vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}$.Since $\vec{r}$ is a unit vector along $\vec{b} + \vec{c}$,we have $\vec{r} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$.Given $\vec{r} \cdot \vec{a} = 3$,so $\frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{b} + \vec{c}|} = 3$.Calculate $(\vec{b} + \vec{c}) \cdot \vec{a} = (5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 5(1) + 2(2) + 3(\lambda-5) = 5 + 4 + 3\lambda - 15 = 3\lambda - 6$.Calculate $|\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda-5)^2} = \sqrt{25 + 4 + \lambda^2 - 10\lambda + 25} = \sqrt{\lambda^2 - 10\lambda + 54}$.Substituting these into the equation: $\frac{3\lambda - 6}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3$.Dividing by $3$: $\frac{\lambda - 2}{\sqrt{\lambda^2 - 10\lambda + 54}} = 1$.Squaring both sides: $(\lambda - 2)^2 = \lambda^2 - 10\lambda + 54$.$\lambda^2 - 4\lambda + 4 = \lambda^2 - 10\lambda + 54$.$6\lambda = 50$.$3\lambda = 25$.

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$,then $3 \lambda$ is equal to :

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