Question

find a vector whose magnitude is 12 and which is perpendicular to each of the vectors, A=2i^+3j^-2k^ and B=6i^+5j^-2k^​

Answer 1

Answer

Given -

$\bf \vec{A} = 2\hat{\imath} + 3\hat{\jmath} - 2\hat{k}$

$\bf \vec{B} = 6\hat{\imath} + 5\hat{\jmath} - 2\hat{k}$

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To find -

A Vector whose magnitude is 12 and is perpendicular to both the vectors $\bf \vec{C}$

$\bf \pink{\vec{C} = \dfrac{ 12 ( \vec{A} \times \vec{B})}{(| \vec{A} \times \vec{B} |) }}$

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Solution -

The cross product of two vectors $\bf \vec{A}$ and $\bf \vec{B}$ gives a vector perpendicular to both the vector.

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Cross Product of A and B = $\bf \vec{A} \times \vec{B} =$

$\left[\begin{array}{ccc}\hat{\imath}&\hat{\jmath}&\hat{k}\\\bf a_x&\bf a_y&\bf a_z\\\bf b_x&\bf b_y& \bf b_z\end{array}\right] =$

$\left[\begin{array}{ccc}\hat{\imath}&\hat{\jmath}&\hat{k}\\\bf 2&\bf 3&\bf -2\\\bf 6&\bf 5& \bf - 2\end{array}\right]$

$\bf \vec{A} \times \vec{B} = [ 3 \times ( - 2) - 5 \times ( - 2)] \hat{\imath} - [2 \times (- 2) - 6 \times ( - 2) ]\hat{\jmath}+ (2 \times 5) -( 6 \times 3)]\hat{k}$

$\bf = ( - 6 + 10)\hat{\imath} - ( - 4 + 12) \hat{\jmath} + (10 - 18) \hat{k}$

$\bf = 4\hat{\imath} - 8\hat{\jmath} - 8\hat{k}$

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$\bf\red{ \vec{A} \times \vec{B} = 4\hat{\imath} - 8\hat{\jmath} - 8\hat{k}}$

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Magnitude of $\bf \vec{A} \times \vec{B}$

$\bf = \sqrt{4^2 + 8 ^2 + 8^2 }$

$\bf = \sqrt{64 + 64 + 16}$

$\bf = \sqrt{144}$

$\bf = 12$

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$\bf \vec{C} = \dfrac{ 12 ( \vec{A} \times \vec{B})}{(| \vec{A} \times \vec{B} |) }$

$\bf = \dfrac{ 12 ( 4\hat{\imath} - 8\hat{\jmath} - 8\hat{k})}{12}$

$\bf = 4\hat{\imath} - 8\hat{\jmath} - 8\hat{k}$

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A Vector whose magnitude is 12 and is perpendicular to both the vectors $\bf\pink{ 4\hat{\imath} - 8\hat{\jmath} - 8\hat{k}}$.

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