{"id":2612,"date":"2021-07-13T19:45:44","date_gmt":"2021-07-13T19:45:44","guid":{"rendered":"https:\/\/mathemerize.com\/?p=2612"},"modified":"2021-10-08T01:28:13","modified_gmt":"2021-10-07T19:58:13","slug":"cross-product-of-vectors-formula","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/cross-product-of-vectors-formula\/","title":{"rendered":"Cross Product of Vectors Formula [ Vector Product ]"},"content":{"rendered":"
Let \\(\\vec{a}\\) & \\(\\vec{b}\\) are two vectors & \\(\\theta\\) is the angle between them, then cross product of vectors formula is,<\/p>\n
\n\\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\) = |\\(\\vec{a}\\)||\\(\\vec{b}\\)|sin\\(\\theta\\)\\(\\hat{n}\\)<\/p>\n
where \\(\\hat{n}\\) is the unit vector perpendicular to both \\(\\vec{a}\\) & \\(\\vec{b}\\).<\/p>\n<\/blockquote>\n
Properties of Vector Cross Product :<\/h2>\n
(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.<\/p>\n
(ii) \\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\) \\(\\ne\\) \\(\\vec{b}\\) \\(\\times\\) \\(\\vec{a}\\) (not commutative)<\/p>\n
(iii) m(\\(\\vec{a}\\)) \\(\\times\\) \\(\\vec{b}\\) = \\(\\vec{a}\\) \\(\\times\\) (m\\(\\vec{b}\\)) = m(\\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\)) where m is a scalar.<\/p>\n
(iv) \\(\\vec{a}\\) \\(\\times\\) (\\(\\vec{b}\\) + \\(\\vec{c}\\)) (distributive over addition)<\/p>\n
(v) \\(\\hat{i}\\) \\(\\times\\) \\(\\hat{i}\\) = \\(\\hat{j}\\) \\(\\times\\) \\(\\hat{j}\\) = \\(\\hat{k}\\) \\(\\times\\) \\(\\hat{k}\\) = 0<\/p>\n
(vi) \\(\\hat{i}\\) \\(\\times\\) \\(\\hat{j}\\) = \\(\\hat{k}\\), \\(\\hat{j}\\) \\(\\times\\) \\(\\hat{k}\\) = \\(\\hat{i}\\), \\(\\hat{k}\\) \\(\\times\\) \\(\\hat{i}\\) = \\(\\hat{j}\\)<\/p>\n
(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}\\)<\/p>\n\n\nExample : <\/span>Find \\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\), if \\(\\vec{a}\\) = \\(2\\hat{i} +\\hat{k}\\) and \\(\\vec{b}\\) = \\(\\hat{i} +\\hat{j} + \\hat{k}\\)<\/p>\n
Solution : <\/span>We have, \\(\\vec{a}\\) = \\(2\\hat{i} +\\hat{k}\\) and \\(\\vec{b}\\) = \\(\\hat{i} +\\hat{j} + \\hat{k}\\)
\n \\(\\therefore\\) \\(\\vec{a}\\times\\vec{b}\\) = \\(\\begin{vmatrix}\n \\hat{i} & \\hat{j} & \\hat{k} \\\\\n 2 & 0 & 1 \\\\\n 1 & 1 & 1 \\\\\n \\end{vmatrix}\\) = \\(-1\\hat{i} – 1\\hat{j} + 2\\hat{k}\\)
\n <\/p>\n\n\nVectors Normal to the Plane of Two Given Vectors :<\/h2>\n
Let \\(\\vec{a}\\) & \\(\\vec{b}\\) be two non-zero, non-parallel vectors and let \\(\\theta\\) be the angle between them.<\/p>\n
\\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\) = |\\(\\vec{a}\\)||\\(\\vec{b}\\)|sin\\(\\theta\\)\\(\\hat{n}\\),<\/p>\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.<\/p>\n
\\(\\therefore\\) \\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\) = | \\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\)|\\(\\hat{n}\\)<\/p>\n
\\(\\implies\\) \\(\\hat{n}\\) = \\(\\vec{a}\\times\\vec{b}\\over |\\vec{a}\\times\\vec{b}|\\)<\/p>\n
\nThus, \\(\\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}\\).<\/p>\n
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}\\).<\/p>\n
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}|\\).<\/p>\n<\/blockquote>\n\n\n
Example : <\/span>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}\\).<\/p>\n
Solution : <\/span>Let \\(\\vec{a}\\) = \\(\\hat{i} -2\\hat{j} + 3\\hat{k}\\) and \\(\\vec{b}\\) = \\(\\hat{i} + 2\\hat{j} – \\hat{k}\\)
\n\\(\\therefore\\) \\(\\vec{a}\\times\\vec{b}\\) = \\(\\begin{vmatrix}\n \\hat{i} & \\hat{j} & \\hat{k} \\\\\n 1 & -2 & 3 \\\\\n 1 & 2 & -1 \\\\\n \\end{vmatrix}\\) = \\((2-6)\\hat{i} – (-1-3)\\hat{j} + (2+2)\\hat{k}\\)
\n= \\(-4\\hat{i} + 4\\hat{j} + 4\\hat{k}\\)
\n\\(\\therefore\\) | \\(\\vec{a}\\) \\(\\times\\) \\(\\vec{b}\\)| = \\(\\sqrt{(-4)^2+4^2+4^2}\\) = 4\\(\\sqrt{3}\\)
\nHence, a unit vector perpendicular to vectors \\(\\vec{a}\\) and \\(\\vec{b}\\) is given by
\n\\(\\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}\\)).\n <\/p>\n\n\nHope you learnt cross product of vectors formula, learn more concepts of vectors and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\n