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Find a Vector of Magnitude √ 171 Which is Perpendicular to Both of the Vectors → a = ^ I + 2 ^ J − 3 ^ K and → a = ^ I + 2 ^ J − 3 ^ K . - Mathematics

Short Note

Find a vector of magnitude \[\sqrt{171}\]  which is perpendicular to both of the vectors \[\vec{a} = \hat{ i } + 2 \hat{ j }  - 3 \hat{ k } \]  and  \[\vec{a} = \hat{ i } + 2 \hat{ j }  - 3 \hat{ k } \] . 

 
 
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Solution

The given vectors are \[\vec{a} = \hat{ i }  + 2 \hat{ j }  - 3 \hat{ k } \] \text{ and }  \[\vec{b} = 3 \hat{ i }  - \hat{ j }  + 2 \hat{ k } \] Unit vectors perpendicular to both \[\vec{a}\]  and \[\vec{b}\] = \[\pm \frac{\vec{a} \times \vec{b}}{\left| \vec{a} \times \vec{b} \right|}\]

Now,

\[\vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i }  & \hat{ j }  & \hat{ k }  \\ 1 & 2 & - 3 \\ 3 & - 1 & 2\end{vmatrix} = \hat{ i } - 11 \hat{ j }  - 7 \hat{ k }  \]

\[ \therefore \left| \vec{a} \times \vec{b} \right| = \left| \hat{ i }  - 11 \hat{ j }  - 7 \hat{ k  } \right| = \sqrt{1^2 + \left( - 11 \right)^2 + \left( - 7 \right)^2} = \sqrt{1 + 121 + 49} = \sqrt{171}\]

Unit vectors perpendicular to both \[\vec{a}\] and\[\vec{ b }\]  =  \[\pm \frac{{i }  - 11 \hat{ j }  - 7 \hat{ k }} {\sqrt{171}}\]

∴ Required vectors = \[\sqrt{171}\left( \pm \frac{\hat{ i }  - 11 \hat{ j }  - 7 \hat{ k } }{\sqrt{171}} \right) = \pm \left( \hat{ i  } - 11 \hat{ j } - 7 \hat{ k }  \right)\]
 
Thus, the vectors of magnitude \[\sqrt{171}\]  which are perpendicular to both the given vectors are \[\pm \left( \hat{ i }  - 11 \hat{ j }  - 7 \hat{ k } \right)\] .
 
 
 
 
 
 
  
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
very short answers | Q 29 | Page 34
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