Let vec{a}=2 hat{i}-3 hat{j}-5 hat{k} and vec | vedclass

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(D) Let $\vec{c} = \vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & -5 \ 3 & 2 & -5 \end{vmatrix} = (15 - (-10))\hat{

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$46$

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(D) Let $\vec{c} = \vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & -5 \ 3 & 2 & -5 \end{vmatrix} = (15 - (-10))\hat{i} - (-10 - (-15))\hat{j} + (4 - (-9))\hat{k} = 25\hat{i} - 5\hat{j} + 13\hat{k}$.Since $\vec{r}$ lies in the plane of $\vec{a}$ and $\vec{b}$,$\vec{r}$ is perpendicular to $\vec{c}$.Also,$\vec{r}$ is orthogonal to $\vec{d} = 5\hat{i} - 2\hat{j} + 3\hat{k}$.Thus,$\vec{r}$ is parallel to $\vec{c} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 25 & -5 & 13 \ 5 & -2 & 3 \end{vmatrix} = (-15 - (-26))\hat{i} - (75 - 65)\hat{j} + (-50 - (-25))\hat{k} = 11\hat{i} - 10\hat{j} - 25\hat{k}$.Let $\vec{r} = \lambda(11\hat{i} - 10\hat{j} - 25\hat{k})$.Given $|\vec{r}| = \sqrt{94}$,we have $|\lambda| \sqrt{11^2 + (-10)^2 + (-25)^2} = \sqrt{94}$.$|\lambda| \sqrt{121 + 100 + 625} = \sqrt{846} = 3\sqrt{94} = \sqrt{94}$.So,$|\lambda| = \frac{1}{3}$.Then $|\vec{r} \cdot \vec{b}| = |\lambda| |(11\hat{i} - 10\hat{j} - 25\hat{k}) \cdot (3\hat{i} + 2\hat{j} - 5\hat{k})| = \frac{1}{3} |33 - 20 + 125| = \frac{1}{3} |138| = 46$.

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