question_answer

Find the equation of the plane passing through the intersection of the planes

\[\vec{r}\cdot (\hat{i}+3\hat{j})-6=0\] and \[\vec{r}\cdot (3\hat{i}-\hat{j}-4\hat{k})=0,\] whose perpendicular distance from origin is unity.

OR

If  \[\vec{a}\cdot \vec{b}=\vec{a}\cdot \vec{c},\,\,\vec{a}\times \vec{b}=\vec{a}\times \vec{c}\] and \[\vec{a}\ne \vec{0},\] then prove that \[\vec{b}=\vec{c}.\]

Answer:

Given equations of planes are \[\vec{r}\cdot (\hat{i}+3\hat{j})-6=0\] and  \[\vec{r}\cdot (3\hat{i}-\hat{j}-4\hat{k})=0.\]

On comparing with \[\vec{r}\cdot \vec{n}=d,\] we get

\[{{\vec{n}}_{1}}=(\hat{i}+3\hat{j}),\] \[{{d}_{1}}=6\] and \[{{\vec{n}}_{2}}=(3\hat{i}-\hat{j}-4\hat{k}),\]

\[{{d}_{2}}=0\]

Now, equation of the plane through the intersection of given planes is

\[\vec{r}\cdot ({{\vec{n}}_{1}}+\lambda {{\vec{n}}_{2}})={{d}_{1}}+{{d}_{2}}\lambda \]

\[\Rightarrow \]   \[\vec{r}\cdot [(\hat{i}+3\hat{j})+\lambda (3\hat{i}-\hat{j}-4\hat{k})]=6+0\cdot \lambda \]

\[\Rightarrow \]   \[\vec{r}\cdot [(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}\,\,+(-\,4\lambda )\hat{k}]=6\]?(i)

On dividing both sides by

\[\sqrt{{{(1+3\lambda )}^{2}}+{{(3-\lambda )}^{2}}+{{(-\,4\lambda )}^{2}},}\] we get

\[\frac{\vec{r}\cdot [(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}\,\,+(-\,4\lambda )\hat{k}]}{\sqrt{{{(1+3\lambda )}^{2}}+{{(3-\lambda )}^{2}}+{{(-\,4\lambda )}^{2}}}}\]

\[=\frac{6}{\sqrt{{{(1+3\lambda )}^{2}}+{{(3-\lambda )}^{2}}+{{(-\,4\lambda )}^{2}}}}\]

According to the question, the perpendicular distance of required plane from origin is unity.

\[\therefore \]      \[=\frac{6}{\sqrt{{{(1+3\lambda )}^{2}}+{{(3-\lambda )}^{2}}+{{(-\,4\lambda )}^{2}}}}=1\]

\[\Rightarrow \]   \[\sqrt{{{(1+3\lambda )}^{2}}+{{(3-\lambda )}^{2}}+{{(-\,4\lambda )}^{2}}}=36\]

[on squaring both side]

\[\Rightarrow \]   \[1+9{{\lambda }^{2}}+6\lambda +9+{{\lambda }^{2}}-6\lambda +16{{\lambda }^{2}}=36\]

\[\begin{align}   & [\because \,\,\,{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\,\,\,\text{and} \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab] \\ \end{align}\]

\[\Rightarrow \]   \[26{{\lambda }^{2}}+10=36\]

\[\Rightarrow \]   \[{{\lambda }^{2}}=1\]

\[\Rightarrow \]   \[\lambda =\pm 1\]

On putting the values of \[\lambda \] in Eq. (i), we get

\[\vec{r}\cdot [(1\pm 3)\hat{i}\,+(3\mp 1)\hat{j}\,+(\mp \,4)\hat{k}]=6\]

\[\Rightarrow \]   \[\vec{r}\cdot [(1+3)\hat{i}\,+(3-1)\hat{j}\,+(-\,4)\hat{k}]=6\]

or         \[\vec{r}\cdot [(1-3)\hat{i}\,+(3+1)\hat{j}\,+4\hat{k}]=6\]

\[\Rightarrow \]   \[\vec{r}\cdot (4\hat{i}\,+2\hat{j}\,-4\hat{k})=6\]

or         \[\vec{r}\cdot (-\,2\hat{i}\,+4\hat{j}\,+4\hat{k})=6\]

\[\Rightarrow \]   \[4x+2y-4z-6=0\]

or         \[-\,2x+4y+4z-6=0\]

OR

Given    \[\vec{a}\cdot \vec{b}=\vec{a}\cdot \vec{c}\] and \[\vec{a}\ne \vec{0}\]

\[\Rightarrow \]   \[\vec{a}\cdot \vec{b}\,-\vec{a}\cdot \vec{c}=\vec{0}\] and \[\vec{a}\ne \vec{0}\]

\[\Rightarrow \]   \[\vec{a}\cdot (\vec{b}\,-\vec{c})=\vec{0}\] and \[\vec{a}\ne \vec{0}\]

\[\Rightarrow \]   \[\vec{b}\,-\vec{c}=0\] or \[\vec{a}\bot (\vec{b}\,-\vec{c})\]

\[\Rightarrow \]   \[\vec{b}=\vec{c}\] or \[\vec{a}\bot (\vec{b}-\vec{c})\]                    ?(i)

Also, given \[\vec{a}\times \vec{b}=\vec{a}\times \vec{c}=\vec{0}\] and \[\vec{a}\ne \vec{0}\]

\[\Rightarrow \]   \[\vec{a}\times \vec{b}-\vec{a}\times \vec{c}=\vec{0}\] and \[\vec{a}\ne \vec{0}\]

\[\Rightarrow \]   \[\vec{a}\times (\vec{b}-\vec{c})=\vec{0}\] and \[\vec{a}\ne \vec{0}\]

\[\Rightarrow \]   \[\vec{b}\,-\vec{c}=\vec{0}\] or \[\vec{a}||(\vec{b}-\vec{c})\]

\[\Rightarrow \]   \[\vec{b}=\vec{c}\] or \[\vec{a}||(\vec{b}-\vec{c})\]                      ?(ii)

From Eqs. (i) and (ii), we get

\[\vec{b}=\vec{c}\]        [\[\because \]\[\vec{a}\bot (\vec{b}-\vec{c})\] and \[\vec{a}||(\vec{b}-\vec{c})\]                        cannot hold simultaneously]

Hence proved