$If\ \overrightarrow{\mathbf{\alpha}} = x\left( \overrightarrow{\mathbf{a}} imes \overrightarrow{\mathbf{b}} ight) + y\left( \overrightarrow{\mathbf{b}} imes \overrightarrow{\mathbf{b}} ight) + z\left( \overrightarrow{\mathbf{c}} imes \overrightarrow{\mathbf{a}} ight)\$and$\ \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ight brack = \frac{1}{8},\$then x + y + z is equal to

Question

$If\ \overrightarrow{\mathbf{\alpha}} = x\left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight) + y\left( \overrightarrow{\mathbf{b}} 	imes \overrightarrow{\mathbf{b}} ight) + z\left( \overrightarrow{\mathbf{c}} 	imes \overrightarrow{\mathbf{a}} ight)\$and$\ \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack = \frac{1}{8},\$then x + y + z is equal to

Dr. Abubacker · Accepted Answer

(a)
We have ,$\overrightarrow{\alpha} = x\left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight) + y\left( \overrightarrow{\mathbf{b}} 	imes \overrightarrow{\mathbf{c}} ight) + z(\overrightarrow{\mathbf{c}} 	imes \overrightarrow{\mathbf{a}})$
Taking dot product with $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ respectively, we get
$\overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{a}} = y\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack\ \  \Longrightarrow y = 8\left( \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{a}} ight)$
$\overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{b}} = z\left( \left( \overrightarrow{\mathbf{c}} 	imes \overrightarrow{\mathbf{a}} ight) \bullet \overrightarrow{\mathbf{b}} ight)$
$\Longrightarrow \ \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{b}} = z\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack\  \Longrightarrow z = 8\left( \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{b}} ight)$
$and\ \ \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{c}} = x\left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}}.\ \overrightarrow{\mathbf{c}} ight)$
$\ \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{c}} = x\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack \Longrightarrow x = 8\left( \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{c}} ight)$
$	herefore x + y + z = 8\overrightarrow{\alpha} \bullet (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}})$

Explanation

Reviewed by Dr. Abubacker

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