Let a=2i+j-k,b=((a (i+j)) i) i.

Question

Let $\vec{a}=2\hat{i}+\hat{j}-\hat{k},\vec{b}=((\vec{a}	imes (\hat{i}+\hat{j}))	imes \hat{i})	imes \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

Accepted Answer

$\begin{gathered} \vec{a}=2\hat{i}+\hat{j}-\hat{k} \ \vec{b}=((\vec{a}	imes (\hat{i}+\hat{j}))	imes \hat{i})	imes \hat{i} \ \vec{a}	imes (\hat{i}+\hat{j})=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 2 \quad  1 \quad  -1 \ 1 \quad  1 \quad  0\end{array}| \ =\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \ =\hat{i}-\hat{j}+\hat{k} \end{gathered}$

$\begin{gathered} (\vec{a}	imes (\hat{i}+\hat{j}))	imes \hat{i}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  -1 \quad  1 \ 1 \quad  0 \quad  0\end{array}| \ \, =\hat{i}(0)-\hat{j}(-1)+\hat{k}(1) \ \, =\hat{j}+\hat{k} \end{gathered}$

$((\overset{	o }{a}	imes (\hat{i}+\hat{j})	imes \hat{i})	imes \hat{i}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 0 \quad  1 \quad  1 \ 1 \quad  0 \quad  0 \end{vmatrix}$

$\begin{gathered} =\hat{i}(0)-\hat{j}(-1)+\hat{k}(-1) \ \vec{b}= \hat{j}-\hat{k} \end{gathered}$

Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$

Square of projection $=2$

Solutions · 1