If bar{a}=hat{i}+hat{j}+hat{k} and bar{b}=hat | vedclass

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(A) Let $\bar{c} = x\hat{i} + y\hat{j} + z\hat{k}$.Given $\bar{a} \cdot \bar{c} = 3$,we have $x + y + z = 3$ (Equation $1$).Given $\bar{a} 	imes \bar

Vedclass · Accepted Answer

$\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}$

Vedclass · Answer

(A) Let $\bar{c} = x\hat{i} + y\hat{j} + z\hat{k}$.Given $\bar{a} \cdot \bar{c} = 3$,we have $x + y + z = 3$ (Equation $1$).Given $\bar{a} 	imes \bar{c} = \bar{b}$,we calculate the cross product:$\bar{a} 	imes \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = (z-y)\hat{i} - (z-x)\hat{j} + (y-x)\hat{k}$.Equating this to $\bar{b} = 0\hat{i} + 1\hat{j} - 1\hat{k}$,we get:$z - y = 0 \implies z = y$$-(z - x) = 1 \implies x - z = 1 \implies x = z + 1$$y - x = -1$Substituting $z = y$ and $x = z + 1$ into Equation $1$:$(z + 1) + z + z = 3 \implies 3z + 1 = 3 \implies 3z = 2 \implies z = \frac{2}{3}$.Then $y = \frac{2}{3}$ and $x = \frac{2}{3} + 1 = \frac{5}{3}$.Thus,$\bar{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.The correct option is $A$.

If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=\hat{j}-\hat{k}$,find a vector $\bar{c}$ such that $\bar{a} \times \bar{c}=\bar{b}$ and $\bar{a} \cdot \bar{c}=3$.

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