Question

Find a unit vector perpendicular to both the vectors \[\vec{a} + \vec{b} \text { and } \vec{a} - \vec{b}\] ,where \[\vec{a} = \hat{i}+ \hat{j} + \hat{k} , \vec{b} =\hat {i} + 2 \hat{j} + 3 \hat{k}\].

Answer 1

Let the unit vector be \[\vec{r}\] = \[x \hat{i} + y \hat{j} + z \hat{k}\].

\[\Rightarrow \sqrt{x^2 + y^2 + z^2} = 1\]

\[ \Rightarrow x^2 + y^2 + z^2 =   1 . . . \left( 1 \right)\]

\[\vec{a} + \vec{b} = 2\hat {i} + 3 \hat{j} + 4 \hat{k} \] \[\text { and } \] \[ \vec{a} - \vec{b} = -\hat{ j } - 2 \hat{k} \]

Since 

\[\vec{r}\] is perpendicular to  \[\vec{a} + \vec{b}\] and \[\vec{a} - \vec{b}\] ,

\[\vec{r} . \left( \vec{a} + \vec{b} \right) = 0\] \[\text { and } \vec{r} . \left( \vec{a} - \vec{b} \right) = 0\]

\[\left( x\hat{ i} + y \hat{j} + z \hat{k} \right) . \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 0\]

\[ \Rightarrow 2x + 3y + 4z = 0 . . . \left( 2 \right)\]

\[\text { and } \left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( - \hat{j} - 2 \hat {k} \right) = 0\]

\[ \Rightarrow - y - 2z = 0 . . . \left( 3 \right)\]

\[\Rightarrow y = - 2z\]

\[\text { Putting the value of y in equation } \left( 2 \right), \text { we get }: \]

\[2x + 3\left( - 2z \right) + 4z = 0\]

\[ \Rightarrow x = z\]

\[\text { Substituting the values of x and y in equation }\left( 1 \right), \text { we get }:\]

\[z^2 + 4 z^2 + z^2 = 1\]

\[z = \pm \frac{1}{\sqrt{6}}\]

\[ \Rightarrow x = \pm \frac{1}{\sqrt{6}}\]

\[ \Rightarrow y = \mp \frac{2}{\sqrt{6}}\]

Hence the vectors are 

\[\frac{1}{\sqrt{6}}\hat{ i } - \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k} \]

\[\text { and } \]\[ - \frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \]