Question 198: Motion in 2D & Vectors

Question

If $$\left| {\overrightarrow A  	imes \overrightarrow B } ight| = \sqrt 3 \overrightarrow A .\overrightarrow B $$ then the value of $$\left| {\overrightarrow A  + \overrightarrow B } ight|$$ is

Accepted Answer

$$\left| {\overrightarrow A  	imes \overrightarrow B } ight| = AB\sin 	heta $$
$$\overrightarrow A \overrightarrow B  = AB\cos 	heta $$
$$\left| {\overrightarrow A  	imes \overrightarrow B } ight| = \sqrt 3 \overrightarrow A \overrightarrow B $$
$$ \Rightarrow AB\sin 	heta  = \sqrt 3 AB\cos 	heta $$
$$ \Rightarrow 	an 	heta  = \sqrt 3 $$    $$ 	herefore 	heta $$ = 60
o
$$ 	herefore $$ $$\left| {\overrightarrow A  	imes \overrightarrow B } ight| = \sqrt {{A^2} + {B^2} + 2AB\cos 60^\circ } $$
$$ = \sqrt {{A^2} + {B^2} + AB} $$ = (A
2
+ b
2
+ AB)
1/2

Answer

(A2  + b2 + AB)1/2

Answer

$${\left( {{A^2} + {B^2} + {{AB} \over {\sqrt 3 }}} \right)^{1/2}}$$

Answer

A + B

Answer

(A2 + B2 + $${\sqrt 3 }$$AB)1/2.

Motion in 2D & Vectors

If $\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \overrightarrow A .\overrightarrow B then the value of \left| {\overrightarrow A + \overrightarrow B } \right|$ is

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