Question

find the cross product of the two vectors
A = 9i + 6j - 8k
B = 2i - 4j + 9k

These i, j and k are i cap, j cap and k cap respectively.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given vectors are

$\rm :\longmapsto\:\vec{a} = 9\hat{i} + 6\hat{j} - 8\hat{k}$

and

$\rm :\longmapsto\:\vec{b} = 2\hat{i} - 4\hat{j} + 9\hat{k}$

So,

$\red{\rm :\longmapsto\:\vec{a} \times \vec{b}}$

$\rm \: = \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\9&6& - 8\\2& - 4& 9\end{array}\right | \end{gathered}$

$\rm \:  =  \:\hat{i}\begin{array}{|cc|}\sf 6 &\sf - 8 \\ \sf - 4 &\sf 9 \\\end{array} - \hat{j}\begin{array}{|cc|}\sf 9 &\sf - 8 \\ \sf 2 &\sf 9 \\\end{array} + \hat{k}\begin{array}{|cc|}\sf 9 &\sf 6 \\ \sf 2 &\sf - 4 \\\end{array}$

$\rm \:  =  \:\hat{i}(54 - 32) - \hat{j}(81 + 16) + \hat{k}( - 36 - 12)$

$\rm \:  =  \:22\hat{i} - 97\hat{j} - 48\hat{k}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \vec{a} \times \vec{b} =  \:22\hat{i} - 97\hat{j} - 48\hat{k} \: }}$

More to know :-

$\boxed{ \tt{ \: \vec{a} \times \vec{a} = 0 \: }}$

$\boxed{ \tt{ \: \vec{a} \times \vec{b} \: = \: - \: \vec{b} \times \vec{a} \: }}$

$\boxed{ \tt{ \: |\vec{a} \times \vec{b}| = |\vec{a}| \: |\vec{b}| \: sin \theta \: }}$

$\boxed{ \tt{ \: \vec{a} \times \vec{b} = 0 \: \: \rm \implies\:\vec{a} \: \parallel \: \vec{b} \: }}$

$\boxed{ \tt{ \: { |\vec{a} \times \vec{b}| }^{2} + { |\vec{a}.\vec{b}| }^{2} = { |\vec{a}| }^{2} \: { |\vec{b}| }^{2} \: }}$