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Question
Find the values of x and y if the vectors \[\vec{a} = 3 \hat{i} + x \hat{j} - \hat{k} \text{ and } \vec{b} = 2 \hat{i} + \hat{j} + y \hat{k}\] are mutually perpendicular vectors of equal magnitude.
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Solution
\[\text{ We have }\]
\[ \vec{a} = 3 \hat{i} + x \hat{j} - \hat{k} \text{ and } \vec{b} = 2 \hat{i} + \hat{j} + y \hat{k} \]
\[\text{It is given that the vectors are perpendicular}.\]
\[ \Rightarrow \vec{a} . \vec{b} = 0\]
\[ \Rightarrow 6 + x - y = 0\]
\[ \Rightarrow x - y = - 6 . . . \left( 1 \right)\]
\[\text{ Also, it is given that }\]
\[\left| \vec{a} \right| = \left| \vec{b} \right|\]
\[ \Rightarrow \sqrt{9 + x^2 + 1} = \sqrt{4 + 1 + y^2}\]
\[ \Rightarrow \sqrt{10 + x^2} = \sqrt{5 + y^2}\]
\[ \Rightarrow 10 + x^2 = 5 + y^2 \]
\[ \Rightarrow x^2 - y^2 = - 5\]
\[ \Rightarrow \left( x + y \right)\left( x - y \right) = - 5\]
\[ \Rightarrow - 6 \left( x + y \right) = - 5 .........................\left[\text{ Using } \left( 1 \right) \right]\]
\[ \Rightarrow x + y = \frac{5}{6} . . . \left( 2 \right)\]
\[(1)+(2) \text{ gives }\]
\[2x = \frac{- 31}{6}\]
\[ \Rightarrow x = \frac{- 31}{12}\]
\[\text{ From } (1),\]
\[\frac{- 31}{12} - y = - 6\]
\[ \Rightarrow y = \frac{- 31}{12} + 6 = \frac{41}{12}\]
\[ \therefore x = \frac{- 31}{12} \text{ and } y = \frac{41}{12}\]
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