Let vec{a}=4 hat{i}+3 hat{j} and vec{b}=3 hat | vedclass

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(A) Given $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$.First,calculate $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i}

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$\frac{5}{\sqrt{2}}$

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(A) Given $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$.First,calculate $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 3 & 0 \ 3 & -4 & 5 \end{vmatrix} = \hat{i}(15-0) - \hat{j}(20-0) + \hat{k}(-16-9) = 15 \hat{i} - 20 \hat{j} - 25 \hat{k}$.Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$.From $\vec{c} \cdot (\vec{a} 	imes \vec{b}) + 25 = 0$,we get $15x - 20y - 25z = -25$,which simplifies to $3x - 4y - 5z = -5$.From $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$,we get $x + y + z = 4$.From the projection of $\vec{c}$ on $\vec{a}$ being $1$,$\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1 \Rightarrow \frac{4x + 3y}{5} = 1 \Rightarrow 4x + 3y = 5$.Solving the system of equations:$1) 3x - 4y - 5z = -5$$2) x + y + z = 4 \Rightarrow 5x + 5y + 5z = 20$Adding $(1)$ and $(2)$: $8x + y = 15 \Rightarrow y = 15 - 8x$.Substitute into $4x + 3y = 5$: $4x + 3(15 - 8x) = 5 \Rightarrow 4x + 45 - 24x = 5 \Rightarrow -20x = -40 \Rightarrow x = 2$.Then $y = 15 - 8(2) = -1$,and $z = 4 - 2 - (-1) = 3$.So,$\vec{c} = 2 \hat{i} - \hat{j} + 3 \hat{k}$.The projection of $\vec{c}$ on $\vec{b}$ is $\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|} = \frac{(2)(3) + (-1)(-4) + (3)(5)}{\sqrt{3^2 + (-4)^2 + 5^2}} = \frac{6 + 4 + 15}{\sqrt{9 + 16 + 25}} = \frac{25}{\sqrt{50}} = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$.

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