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

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(C) Given $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$.First,calculate $\overrightarrow{b}=\overrightarrow{a}	imes(\hat{i}-2\hat{k})$:$\overrightar

Vedclass · Accepted Answer

$2\sqrt{14}$

Vedclass · Answer

(C) Given $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$.First,calculate $\overrightarrow{b}=\overrightarrow{a}	imes(\hat{i}-2\hat{k})$:$\overrightarrow{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 2 \ 1 & 0 & -2\end{vmatrix} = \hat{i}(2-0) - \hat{j}(-6-2) + \hat{k}(0-(-1)) = 2\hat{i}+8\hat{j}+\hat{k}$.Next,calculate $\overrightarrow{c}=\overrightarrow{b}	imes\hat{k}$:$\overrightarrow{c}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ 2 & 8 & 1 \ 0 & 0 & 1\end{vmatrix} = \hat{i}(8-0) - \hat{j}(2-0) + \hat{k}(0-0) = 8\hat{i}-2\hat{j}$.Now,find $\overrightarrow{c}-2\hat{j} = (8\hat{i}-2\hat{j}) - 2\hat{j} = 8\hat{i}-4\hat{j}$.The projection of a vector $\overrightarrow{v}$ on $\overrightarrow{a}$ is given by $\frac{\overrightarrow{v} \cdot \overrightarrow{a}}{|\overrightarrow{a}|}$.Here,$\overrightarrow{v} = 8\hat{i}-4\hat{j}$ and $\overrightarrow{a} = 3\hat{i}-\hat{j}+2\hat{k}$.$|\overrightarrow{a}| = \sqrt{3^2+(-1)^2+2^2} = \sqrt{9+1+4} = \sqrt{14}$.Projection $= \frac{(8\hat{i}-4\hat{j}) \cdot (3\hat{i}-\hat{j}+2\hat{k})}{\sqrt{14}} = \frac{(8)(3) + (-4)(-1) + (0)(2)}{\sqrt{14}} = \frac{24+4}{\sqrt{14}} = \frac{28}{\sqrt{14}} = 2\sqrt{14}$.

Let $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$ and $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$. Then the projection of $\overrightarrow{c}-2\hat{j}$ on $\overrightarrow{a}$ is:

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