# 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  $\vec{a} = \hat{ i } + 2 \hat{ j } - 3 \hat{ k }$ .

#### 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