Let overrightarrow{a} = alpha hat{i} + 3 hat{ | vedclass

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Question

(A) The projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is given by $\frac{\overrightarrow{a} \cdot \overrightarrow{c}}{|\overrightarrow{c}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is given by $\frac{\overrightarrow{a} \cdot \overrightarrow{c}}{|\overrightarrow{c}|} = \frac{10}{3}$.Calculating $\overrightarrow{a} \cdot \overrightarrow{c} = (\alpha)(1) + (3)(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8$.Calculating $|\overrightarrow{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Thus,$\frac{\alpha + 8}{3} = \frac{10}{3} \Rightarrow \alpha + 8 = 10 \Rightarrow \alpha = 2$.Next,we calculate $\overrightarrow{b} 	imes \overrightarrow{c}$:$\overrightarrow{b} 	imes \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -\beta & 4 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.Given $\overrightarrow{b} 	imes \overrightarrow{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,comparing the components:$2\beta - 8 = -6 \Rightarrow 2\beta = 2 \Rightarrow \beta = 1$.Therefore,$\alpha + \beta = 2 + 1 = 3$.

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