If $$\overrightarrow {\text{a}} $$ and $$\overrightarrow {\text{b}} $$ are two arbitrary vectors with magnitudes a and b, respectively, $${\left| {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right|^2}$$ will be equal to
A. $${{\text{a}}^2}{{\text{b}}^2} - {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$
B. $${\text{ab}} - \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$
C. $${{\text{a}}^2}{{\text{b}}^2} + {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$
D. $${\text{ab}} + \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$
Answer: Option A
This Question Belongs to Engineering Maths >> Calculus
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
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