Trong không gian với hệ tọa độ Oxyz, cho hai vectơ a và b thỏa mãn vecto a = 2,b = 3
a) | S | b) | S | c) | Đ | d) | S |
a. Sai.\[\overrightarrow a \overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right)\]\[ = 2.3.\cos {60^0}\]\[ = 3\]
b. Đúng. \({\left| {\overrightarrow a + \overrightarrow b } \right|^2} = {\overrightarrow a ^2} + 2\overrightarrow {a.} \overrightarrow b + {\overrightarrow b ^2}\)
\( = {\overrightarrow a ^2} + 2\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right) + {\overrightarrow b ^2}\)
\( = {2^2} + 2.2.3.\cos 60^\circ + {3^2} = 19\)
\( \Rightarrow \left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {19} \).
c. Sai. \({\left| {\overrightarrow a - \overrightarrow b } \right|^2} = {\overrightarrow a ^2} - 2\overrightarrow {a.} \overrightarrow b + {\overrightarrow b ^2}\)
\( = {\overrightarrow a ^2} - 2\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right) + {\overrightarrow b ^2}\)
\( = {2^2} - 2.2.3.\cos 60^\circ + {3^2} = 7\)
\( \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right| = \sqrt 7 \).
d. Sai. \({\left| {\overrightarrow a - 2\overrightarrow b } \right|^2} = {\overrightarrow a ^2} - 4\overrightarrow {a.} \overrightarrow b + 4{\overrightarrow b ^2}\)
\( = {\overrightarrow a ^2} - 4\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right) + 4{\overrightarrow b ^2}\)
\( = {2^2} - 4.2.3.\cos 60^\circ + {4.3^2} = 28\)
\( \Rightarrow \left| {\overrightarrow a - 2\overrightarrow b } \right| = \sqrt {28} .\)