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.
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