question_answer

If \[\overset{\to }{\mathop{\text{A}}}\,\times \overset{\to }{\mathop{\text{B}}}\,=\overset{\to }{\mathop{\text{A}}}\,\times \overset{\to }{\mathop{\text{C}}}\,\], is it correct to conclude that \[\overset{\to }{\mathop{\text{B}}}\,=\overset{\to }{\mathop{\text{C}}}\,\]?

Answer:

                Let \[{{\text{ }\!\!\theta\!\!\text{ }}_{1}}\]be the smaller angle between \[\overset{\to }{\mathop{\text{A}}}\,\]and \[\overset{\to }{\mathop{\text{B}}}\,\]; and \[{{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\]be the smaller angle between\[\overset{\to }{\mathop{\text{A}}}\,\] and \[\overset{\to }{\mathop{\text{C}}}\,\].                 Given, \[\overset{\to }{\mathop{\text{A}}}\,\times \overset{\to }{\mathop{\text{B}}}\,=\overset{\to }{\mathop{\text{A}}}\,\times \overset{\to }{\mathop{\text{C}}}\,\]                 \[\therefore \]  \[AB\sin {{\text{ }\!\!\theta\!\!\text{ }}_{\text{1}}}{{\hat{n}}_{1}}=AC\sin {{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}{{\hat{n}}_{2}}\] Where\[{{\hat{n}}_{1}}\] and\[{{\hat{n}}_{2}}\] are unit vectors in the direction of vectors\[(\overset{\to }{\mathop{\text{A}}}\,\times \overset{\to }{\mathop{\text{B}}}\,)\] and\[(\overset{\to }{\mathop{\text{A}}}\,\times \overset{\to }{\mathop{\text{C}}}\,)\] respectively. From (i), \[B\sin {{\theta }_{1}}{{\hat{n}}_{1}}=C\sin {{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}{{\hat{n}}_{2}}\] If \[{{\text{ }\!\!\theta\!\!\text{ }}_{1}}={{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\]and\[{{\hat{n}}_{1}}={{\hat{n}}_{2}}\] , then B = C and \[\overset{\to }{\mathop{\text{B}}}\,=\overset{\to }{\mathop{\text{C}}}\,\] If \[{{\text{ }\!\!\theta\!\!\text{ }}_{1}}\ne {{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\]and\[{{\hat{n}}_{1}}\ne {{\hat{n}}_{2}}\] , then \[\text{B}\ne \text{C}\] and \[\overset{\to }{\mathop{\text{B}}}\,\ne \overset{\to }{\mathop{\text{C}}}\,\] If \[{{\text{ }\!\!\theta\!\!\text{ }}_{1}}\ne {{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\]and\[{{\hat{n}}_{1}}={{\hat{n}}_{2}}\] , then \[\text{B}\ne \text{C}\] and \[\overset{\to }{\mathop{\text{B}}}\,\ne \overset{\to }{\mathop{\text{C}}}\,\] If \[{{\text{ }\!\!\theta\!\!\text{ }}_{1}}\text{=}{{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\]and\[{{\hat{n}}_{1}}\ne {{\hat{n}}_{2}}\] , then B = C and \[\overset{\to }{\mathop{\text{B}}}\,\ne \overset{\to }{\mathop{\text{C}}}\,\]