Cross Product of Vectors Formula :
Let \(\vec{a}\) & \(\vec{b}\) are two vectors & \(\theta\) is the angle between them, then cross product of vectors formula is,
\(\vec{a}\) \(\times\) \(\vec{b}\) = |\(\vec{a}\)||\(\vec{b}\)|sin\(\theta\)\(\hat{n}\)
where \(\hat{n}\) is the unit vector perpendicular to both \(\vec{a}\) & \(\vec{b}\).
Properties of Vector Cross Product :
(i) \(\vec{a}\) \(\times\) \(\vec{b}\) = \(\vec{0}\) \(\iff\) \(\vec{a}\) & \(\vec{b}\) are parallel(Collinear) (\(\vec{a}\) \(\ne\) 0, \(\vec{b}\) \(\ne\) 0) i.e. \(\vec{a}\) = K\(\vec{b}\), where K is a scalar.
(ii) \(\vec{a}\) \(\times\) \(\vec{b}\) \(\ne\) \(\vec{b}\) \(\times\) \(\vec{a}\) (not commutative)
(iii) m(\(\vec{a}\)) \(\times\) \(\vec{b}\) = \(\vec{a}\) \(\times\) (m\(\vec{b}\)) = m(\(\vec{a}\) \(\times\) \(\vec{b}\)) where m is a scalar.
(iv) \(\vec{a}\) \(\times\) (\(\vec{b}\) + \(\vec{c}\)) (distributive over addition)
(v) \(\hat{i}\) \(\times\) \(\hat{i}\) = \(\hat{j}\) \(\times\) \(\hat{j}\) = \(\hat{k}\) \(\times\) \(\hat{k}\) = 0
(vi) \(\hat{i}\) \(\times\) \(\hat{j}\) = \(\hat{k}\), \(\hat{j}\) \(\times\) \(\hat{k}\) = \(\hat{i}\), \(\hat{k}\) \(\times\) \(\hat{i}\) = \(\hat{j}\)
(vii) If \(\vec{a}\) = \(a_1\hat{i}\) + \(a_2\hat{j}\) + \(a_3\hat{k}\) \(\vec{b}\) = \(b_1\hat{i}\) + \(b_2\hat{j}\) + \(b_3\hat{k}\), then \(\vec{a}\) \(\times\) \(\vec{b}\) = \(\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
\end{vmatrix}\)
Example : Find \(\vec{a}\) \(\times\) \(\vec{b}\), if \(\vec{a}\) = \(2\hat{i} +\hat{k}\) and \(\vec{b}\) = \(\hat{i} +\hat{j} + \hat{k}\)
Solution : We have, \(\vec{a}\) = \(2\hat{i} +\hat{k}\) and \(\vec{b}\) = \(\hat{i} +\hat{j} + \hat{k}\)
\(\therefore\) \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 1 \\
1 & 1 & 1 \\
\end{vmatrix}\) = \(-1\hat{i} – 1\hat{j} + 2\hat{k}\)
Vectors Normal to the Plane of Two Given Vectors :
Let \(\vec{a}\) & \(\vec{b}\) be two non-zero, non-parallel vectors and let \(\theta\) be the angle between them.
\(\vec{a}\) \(\times\) \(\vec{b}\) = |\(\vec{a}\)||\(\vec{b}\)|sin\(\theta\)\(\hat{n}\),
where \(\hat{n}\) is the unit vector perpendicular to plane of \(\vec{a}\) & \(\vec{b}\) such that \(\vec{a}\), \(\vec{b}\), \(\hat{n}\) form a right-handed system.
\(\therefore\) \(\vec{a}\) \(\times\) \(\vec{b}\) = | \(\vec{a}\) \(\times\) \(\vec{b}\)|\(\hat{n}\)
\(\implies\) \(\hat{n}\) = \(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\)
Thus, \(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) is a unit vector perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\).
Note that -\(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) is also a unit vector perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\).
Vector of magnitude ‘\(\lambda\)’ normal to the plane of \(\vec{a}\) and \(\vec{b}\) are given by \(\pm\)\(\lambda(\vec{a}\times\vec{b})\over |\vec{a}\times\vec{b}|\).
Example : Find a unit vector perpendicular to both the vectors \(\hat{i} -2\hat{j} + 3\hat{k}\) and \(\hat{i} + 2\hat{j} – \hat{k}\).
Solution : Let \(\vec{a}\) = \(\hat{i} -2\hat{j} + 3\hat{k}\) and \(\vec{b}\) = \(\hat{i} + 2\hat{j} – \hat{k}\)
\(\therefore\) \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
1 & 2 & -1 \\
\end{vmatrix}\) = \((2-6)\hat{i} – (-1-3)\hat{j} + (2+2)\hat{k}\)
= \(-4\hat{i} + 4\hat{j} + 4\hat{k}\)
\(\therefore\) | \(\vec{a}\) \(\times\) \(\vec{b}\)| = \(\sqrt{(-4)^2+4^2+4^2}\) = 4\(\sqrt{3}\)
Hence, a unit vector perpendicular to vectors \(\vec{a}\) and \(\vec{b}\) is given by
\(\hat{n}\) = \(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) = \(-4\hat{i} + 4\hat{j} + 4\hat{k}\over 4\sqrt{3}\) = \(1\over \sqrt{3}\)(\(-\hat{i}+\hat{j}+\hat{k}\)).
Hope you learnt cross product of vectors formula, learn more concepts of vectors and practice more questions to get ahead in the competition. Good luck!
