Scalar and Vector Examples

SCALAR AND VECTOR EXAMPLES

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 \(\vec{a}\) & \(\vec{b}\) = \(\pm\)\(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\)

\(\therefore\)   \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}\) = \(-5\hat{i} - 5\hat{j} - 5\hat{k}\)

\(\therefore\)   Unit Vectors = \(\pm\) \(-5\hat{i} - 5\hat{j} - 5\hat{k}\over 5\sqrt{3}\)

Hence the required vectors are \(\pm\) \(5\sqrt{3}\over 3\)(\(\hat{i} + \hat{j} + \hat{k}\))



Example 2 : 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 |\(\vec{c}\)| = |\(\vec{a}\)|

Solution : \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\)

\(\implies\)   \(\vec{c}\perp\vec{a}\) , \(\vec{c}\perp\vec{b}\) and \(\vec{a}\perp\vec{b}\), \(\vec{a}\perp\vec{c}\)

\(\implies\)   \(\vec{a}\perp\vec{b}\), \(\vec{b}\perp\vec{c}\) and \(\vec{c}\perp\vec{a}\)

\(\implies\)   \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors.

Again, \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\)

\(\implies\)   |\(\vec{a}\times\vec{b}\)| = |\(\vec{c}\)| and |\(\vec{b}\times\vec{c}\)| = |\(\vec{a}\)|

\(\implies\)   \(|\vec{a}||\vec{b}|sin{\pi\over 2}\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|sin{\pi\over 2}\) = |\(\vec{a}\)|     (\(\because\) \(\vec{a}\perp\vec{b}\) and \(\vec{b}\perp\vec{c}\))

\(\implies\)   \(|\vec{a}||\vec{b}|\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|\) = |\(\vec{a}\)|

\(\implies\)   \({|\vec{b}|}^2\) |\(\vec{c}\)| = |\(\vec{c}\)|

\(\implies\)   \({|\vec{b}|}^2\) = 1

\(\implies\)   \(|\vec{b}|\) = 1

putting in \(|\vec{a}||\vec{b}|\) = |\(\vec{c}\)|

\(\implies\)   \(|\vec{a}|\) = |\(\vec{c}\)|



Example 3 : 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[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]

Solution : We have [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)]

= {(\(\vec{a}\) + \(\vec{b}\))\(\times\)(\(\vec{b}\) + \(\vec{c}\))}.(\(\vec{c}\) + \(\vec{a}\))

= {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{b}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\))       {\(\vec{b}\)\(\times\)\(\vec{b}\) = 0}

= {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\))

= (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{b}\)).\(\vec{a}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{a}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{a}\)

= [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] + 0 + 0 + 0 + 0 + [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)]

= [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] + [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]


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