question_answer

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

OR

If \[\vec{a}\] and \[\vec{b}\] are two non-zero vectors, then prove that \[{{(\vec{a}\times \vec{b})}^{2}}=\left| \begin{matrix}    \vec{a}\cdot \vec{a} & \vec{a}\cdot \vec{b}  \\    \vec{a}\cdot \vec{b} & \vec{b}\cdot \vec{b}  \\ \end{matrix} \right|.\]

Answer:

Given, \[\vec{a}=\hat{i}+\hat{j}+\hat{k}\] and \[\vec{b}=\hat{j}-\hat{k}\]

Let \[\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}\]

Then, \[\vec{a}\times \vec{c}=\vec{b}\] \[\Rightarrow \] \[\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & 1 & 1  \\    x & y & z  \\ \end{matrix} \right|=\hat{j}-\hat{k}\]

\[\Rightarrow \]   \[\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)=\hat{j}-\hat{k}\]

On comparing the coefficients of \[\hat{i},\,\hat{j}\] and \[\hat{k}\]from both sides, we get

\[z-y=0\]                                    ? (i)

\[x-z=1\]                                    ? (ii)

and       \[x-y=1\]                                   ? (iii)

Also \[\vec{a}\cdot \vec{c}=3\] \[\Rightarrow \] \[(\hat{i}+\hat{j}+\hat{k})\cdot (x\hat{i}+y\hat{j}+z\hat{k})=3\]

\[\Rightarrow \]   \[x+y+z=3\]                             ? (iv)

On adding Eqs. (ii) and (iii), we get

\[2x-y-z=2\]                               ? (v)

On adding Eqs. (iv) and (v), we get

\[3x=5\] \[\Rightarrow \] \[x=\frac{5}{3}\]

Then, from Eq. (iii), \[y=\frac{5}{3}-1=\frac{2}{3}\] and from Eq. (i), \[z=\frac{2}{3}\]

Hence, \[\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}=\frac{1}{3}(5\hat{i}+2\hat{j}+2\,\hat{k})\]

OR

Given, \[\vec{a}\] and \[\vec{b}\] are two non-zero vectors.

\[\therefore \]      \[{{(\vec{a}\times \vec{b})}^{2}}=\,\,|\vec{a}\times \vec{b}{{|}^{2}}\]

\[\Rightarrow \]   \[{{(\vec{a}\times \vec{b})}^{2}}={{\left\{ |\vec{a}|\,|\vec{b}|\,\sin \theta  \right\}}^{2}}\]                  \[[\because \,\,|\hat{n}|\,\,=1]\]

\[\Rightarrow \]   \[{{(\vec{a}\times \vec{b})}^{2}}=|\vec{a}{{|}^{2}}\,|\vec{b}{{|}^{2}}\,{{\sin }^{2}}\theta \]

\[\Rightarrow \]   \[{{(\vec{a}\times \vec{b})}^{2}}=\left\{ |\vec{a}{{|}^{2}}\,|\vec{b}{{|}^{2}} \right\}(1-co{{s}^{2}}\theta )\]

\[\Rightarrow \]   \[{{(\vec{a}\times \vec{b})}^{2}}=\,\,|\vec{a}{{|}^{2}}\,|\vec{b}{{|}^{2}}-|\vec{a}{{|}^{2}}\,|\vec{b}{{|}^{2}}\,{{\cos }^{2}}\theta \]

\[\Rightarrow \]   \[{{(\vec{a}\times \vec{b})}^{2}}=(\vec{a}\cdot \vec{a})\,(\vec{b}\cdot \vec{b})-(\vec{a}\cdot \vec{b})\,(\vec{a}\cdot \vec{b})\]

\[[\because \,\,\,\vec{a}\cdot \vec{b}=|\vec{a}|\,|\vec{b}|\,\,\cos \theta ]\]

\[\Rightarrow \]   \[{{(\vec{a}\times \vec{b})}^{2}}=\left| \begin{matrix}    \vec{a}\cdot \vec{a} & \vec{a}\cdot \vec{b}  \\    \vec{a}\cdot \vec{b} & \vec{b}\cdot \vec{b}  \\ \end{matrix} \right|\]   Hence proved.