Let overrightarrow{a}=3 hat{i}+2 hat{j}+hat{k | vedclass

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(B) Given $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$ and $\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$.First,calculate $\overrightarrow{a}+\o

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(B) Given $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$ and $\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$.First,calculate $\overrightarrow{a}+\overrightarrow{b} = 5 \hat{i}+\hat{j}+4 \hat{k}$ and $\overrightarrow{a} 	imes \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 1 \ 2 & -1 & 3 \end{vmatrix} = \hat{i}(6+1) - \hat{j}(9-2) + \hat{k}(-3-4) = 7 \hat{i}-7 \hat{j}-7 \hat{k}$.The equation $(\overrightarrow{a}+\overrightarrow{b}) 	imes \overrightarrow{c} = 2(\overrightarrow{a} 	imes \overrightarrow{b}) + 24 \hat{j}-6 \hat{k}$ becomes:$(5 \hat{i}+\hat{j}+4 \hat{k}) 	imes \overrightarrow{c} = 2(7 \hat{i}-7 \hat{j}-7 \hat{k}) + 24 \hat{j}-6 \hat{k} = 14 \hat{i}+10 \hat{j}-20 \hat{k}$.Let $\overrightarrow{c} = x \hat{i}+y \hat{j}+z \hat{k}$. Then $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & 1 & 4 \ x & y & z \end{vmatrix} = 14 \hat{i}+10 \hat{j}-20 \hat{k}$.This gives $\hat{i}(z-4y) - \hat{j}(5z-4x) + \hat{k}(5y-x) = 14 \hat{i}+10 \hat{j}-20 \hat{k}$.Equating components: $z-4y=14$,$4x-5z=10$,$5y-x=-20$.Also,$(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c} = -3$. Since $\overrightarrow{a}-\overrightarrow{b}+\hat{i} = (3-2+1)\hat{i} + (2-(-1))\hat{j} + (1-3)\hat{k} = 2 \hat{i}+3 \hat{j}-2 \hat{k}$,we have $2x+3y-2z=-3$.Solving the system: From $x=5y+20$,substitute into others to get $x=5, y=-3, z=2$.Thus,$|\overrightarrow{c}|^2 = 5^2 + (-3)^2 + 2^2 = 25+9+4 = 38$.

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