If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are unit vectors, then the vectors $\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} ight) imes (\overrightarrow{\mathbf{a}} imes \overrightarrow{\mathbf{b}})$ is parallel to the vector

Question

If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are unit vectors, then the vectors $\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} ight) 	imes (\overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}})$ is parallel to the vector

Dr. Abubacker · Accepted Answer

(a)
$\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} ight) 	imes (\overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}})$
$= \overrightarrow{\mathbf{a}} 	imes \left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight) + \overrightarrow{\mathbf{b}} 	imes \left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight)$
$= \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} ight)\overrightarrow{\mathbf{a}} - \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{a}} ight)\overrightarrow{\mathbf{b}} + \left( \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{b}} ight)\overrightarrow{\mathbf{a}} - (\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{a}})\overrightarrow{\mathbf{b}}$
$= \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} ight)\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{a}} - (\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{a}})\overrightarrow{\mathbf{b}}$
$= \left( \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} ight)\left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} - 1 ight)$
∴ Given vector is parallel to$\ \left( \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} ight).$

Q. If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are unit vectors, then the vectors $\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right) \times (\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$ is parallel to the vector

Explanation

Reviewed by Dr. Abubacker

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