Example 1 : <\/span>Find the vector of magnitude 5 which are perpendicular to the vectors \\(\\vec{a}\\) = \\(2\\hat{i} + \\hat{j} – 3\\hat{k}\\) and \\(\\vec{b}\\) = \\(\\hat{i} – 2\\hat{j} + \\hat{k}\\).<\/p>\n Solution : <\/span>Unit vectors perpendicular to \\(\\vec{a}\\) & \\(\\vec{b}\\) = \\(\\pm\\)\\(\\vec{a}\\times\\vec{b}\\over |\\vec{a}\\times\\vec{b}|\\) Example 2 : <\/span> If \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) are three non zero vectors such that \\(\\vec{a}\\times\\vec{b}\\) = \\(\\vec{c}\\) and \\(\\vec{b}\\times\\vec{c}\\) = \\(\\vec{a}\\), prove that \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) are mutually at right angles and |\\(\\vec{b}\\)| = 1 and |\\(\\vec{c}\\)| = |\\(\\vec{a}\\)|<\/p>\n Solution : <\/span>\\(\\vec{a}\\times\\vec{b}\\) = \\(\\vec{c}\\) and \\(\\vec{b}\\times\\vec{c}\\) = \\(\\vec{a}\\) Example 3 : <\/span>For any three vectors \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) prove that [\\(\\vec{a}\\) + \\(\\vec{b}\\) \\(\\vec{b}\\) + \\(\\vec{c}\\) \\(\\vec{c}\\) + \\(\\vec{a}\\)] = 2[\\(\\vec{a}\\) \\(\\vec{b}\\) \\(\\vec{c}\\)]<\/p>\n Solution : <\/span>We have [\\(\\vec{a}\\) + \\(\\vec{b}\\) \\(\\vec{b}\\) + \\(\\vec{c}\\) \\(\\vec{c}\\) + \\(\\vec{a}\\)] Practice these given scalar and vector examples to test your knowledge on concepts of scalar and vectors.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some scalar and vector examples for better understanding of scalar and vector concepts. Example 1 : Find the vector of magnitude 5 which are perpendicular to the vectors \\(\\vec{a}\\) = \\(2\\hat{i} + \\hat{j} – 3\\hat{k}\\) and \\(\\vec{b}\\) = \\(\\hat{i} – 2\\hat{j} + \\hat{k}\\). Solution : Unit vectors perpendicular to \\(\\vec{a}\\) & …<\/p>\n
\n \\(\\therefore\\) \\(\\vec{a}\\times\\vec{b}\\) = \\(\\begin{vmatrix}\n \t \\hat{i} & \\hat{j} & \\hat{k} \\\\\n 2 & 1 & -3 \\\\\n 1 & 2 & -2 \\\\\n \\end{vmatrix}\\) = \\(-5\\hat{i} – 5\\hat{j} – 5\\hat{k}\\)
\n \\(\\therefore\\) Unit Vectors = \\(\\pm\\) \\(-5\\hat{i} – 5\\hat{j} – 5\\hat{k}\\over 5\\sqrt{3}\\)
\n Hence the required vectors are \\(\\pm\\) \\(5\\sqrt{3}\\over 3\\)(\\(\\hat{i} + \\hat{j} + \\hat{k}\\))<\/p>\n
\n\n \n
\n \\(\\implies\\) \\(\\vec{c}\\perp\\vec{a}\\) , \\(\\vec{c}\\perp\\vec{b}\\) and \\(\\vec{a}\\perp\\vec{b}\\), \\(\\vec{a}\\perp\\vec{c}\\)
\n \\(\\implies\\) \\(\\vec{a}\\perp\\vec{b}\\), \\(\\vec{b}\\perp\\vec{c}\\) and \\(\\vec{c}\\perp\\vec{a}\\)
\n \\(\\implies\\) \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) are mutually perpendicular vectors.
\n Again, \\(\\vec{a}\\times\\vec{b}\\) = \\(\\vec{c}\\) and \\(\\vec{b}\\times\\vec{c}\\) = \\(\\vec{a}\\)
\n \\(\\implies\\) |\\(\\vec{a}\\times\\vec{b}\\)| = |\\(\\vec{c}\\)| and |\\(\\vec{b}\\times\\vec{c}\\)| = |\\(\\vec{a}\\)|
\n \\(\\implies\\) \\(|\\vec{a}||\\vec{b}|sin{\\pi\\over 2}\\) = |\\(\\vec{c}\\)| and \\(|\\vec{b}||\\vec{c}|sin{\\pi\\over 2}\\) = |\\(\\vec{a}\\)| (\\(\\because\\) \\(\\vec{a}\\perp\\vec{b}\\) and \\(\\vec{b}\\perp\\vec{c}\\))
\n \\(\\implies\\) \\(|\\vec{a}||\\vec{b}|\\) = |\\(\\vec{c}\\)| and \\(|\\vec{b}||\\vec{c}|\\) = |\\(\\vec{a}\\)|
\n \\(\\implies\\) \\({|\\vec{b}|}^2\\) |\\(\\vec{c}\\)| = |\\(\\vec{c}\\)|
\n \\(\\implies\\) \\({|\\vec{b}|}^2\\) = 1
\n \\(\\implies\\) \\(|\\vec{b}|\\) = 1
\n putting in \\(|\\vec{a}||\\vec{b}|\\) = |\\(\\vec{c}\\)|
\n \\(\\implies\\) \\(|\\vec{a}|\\) = |\\(\\vec{c}\\)|\n <\/p>\n
\n\n \n
\n = {(\\(\\vec{a}\\) + \\(\\vec{b}\\))\\(\\times\\)(\\(\\vec{b}\\) + \\(\\vec{c}\\))}.(\\(\\vec{c}\\) + \\(\\vec{a}\\))
\n = {\\(\\vec{a}\\)\\(\\times\\)\\(\\vec{b}\\) + \\(\\vec{a}\\)\\(\\times\\)\\(\\vec{c}\\) + \\(\\vec{b}\\)\\(\\times\\)\\(\\vec{b}\\) + \\(\\vec{b}\\)\\(\\times\\)\\(\\vec{c}\\)}.(\\(\\vec{c}\\) + \\(\\vec{a}\\)) {\\(\\vec{b}\\)\\(\\times\\)\\(\\vec{b}\\) = 0}
\n = {\\(\\vec{a}\\)\\(\\times\\)\\(\\vec{b}\\) + \\(\\vec{a}\\)\\(\\times\\)\\(\\vec{c}\\) + \\(\\vec{b}\\)\\(\\times\\)\\(\\vec{c}\\)}.(\\(\\vec{c}\\) + \\(\\vec{a}\\))
\n = (\\(\\vec{a}\\times\\vec{b}\\)).\\(\\vec{c}\\) + (\\(\\vec{a}\\times\\vec{c}\\)).\\(\\vec{c}\\) + (\\(\\vec{b}\\times\\vec{c}\\)).\\(\\vec{c}\\) + (\\(\\vec{a}\\times\\vec{b}\\)).\\(\\vec{a}\\) + (\\(\\vec{a}\\times\\vec{c}\\)).\\(\\vec{a}\\) + (\\(\\vec{b}\\times\\vec{c}\\)).\\(\\vec{a}\\)
\n = [\\(\\vec{a}\\) \\(\\vec{b}\\) \\(\\vec{c}\\)] + 0 + 0 + 0 + 0 + [\\(\\vec{b}\\) \\(\\vec{c}\\) \\(\\vec{a}\\)]
\n = [\\(\\vec{a}\\) \\(\\vec{b}\\) \\(\\vec{c}\\)] + [\\(\\vec{a}\\) \\(\\vec{b}\\) \\(\\vec{c}\\)] = 2[\\(\\vec{a}\\) \\(\\vec{b}\\) \\(\\vec{c}\\)]\n <\/p>\n
\n