Question

If $a=i+2j+2k,\left|b\right|=5$ and the angle between $aandbis\frac{\pi }{6}$, then the area of the triangle formed by these two vectors are two sides is:

• A. $\frac{15}{4}$
• B. $\frac{15}{2}$
• C. $15$
• D. $\frac{15\sqrt{3}}{2}$
• E. $15\sqrt{3}$

Answer 1

The correct option is A

$\frac{15}{4}$

Finding the area of the triangle formed by these two vectors:

Given, $a=i+2j+2k,\left|b\right|=5$ & angle between $aandbis\frac{\pi }{6}$

Area of triangle is given by,

$$\begin{gathered} \left|a\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2} \\ \left|a\right|=\sqrt{1^2+2^2+2^2}[a_1=1,a_2=2,a_3=2] \\ \left|a\right|=\sqrt{9} \\ \left|a\right|=3 \end{gathered}$$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\left|a\right|\mathbf{\times }\left|b\right|$

Therefore,

Area

$$\begin{gathered} =\frac{1}{2}\times \left|a\right|\times \left|b\right| \\ =\frac{1}{2}\times 3\times 5\times \sin \frac{\pi }{6} \\ =\frac{1}{2}\times 3\times 5\times \frac{1}{2} \\ =\frac{15}{4} \end{gathered}$$

Hence, correct option is $\left(A\right)$