Let $\overrightarrow{\mathbf{a}},\ \overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}$ be three non-coplanar vectors, and let $\overrightarrow{\mathbf{p}},\ \overrightarrow{\mathbf{q}}\ and\ \overrightarrow{\mathbf{r}}$ be vectors defined by the relations $\overrightarrow{\mathbf{p}} = \frac{\overrightarrow{\mathbf{b}} imes \overrightarrow{\mathbf{c}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ight brack}\ ,\ \overrightarrow{\mathbf{q}} = \frac{\overrightarrow{\mathbf{c}} imes \overrightarrow{\mathbf{a}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ight brack}and\ \ \overrightarrow{\mathbf{r}} = \frac{\overrightarrow{\mathbf{a}} imes \overrightarrow{\mathbf{b}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ight brack}\ \$ Then, the value of the expression $\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} ight) \bullet \overrightarrow{\mathbf{p}} + \left( \overrightarrow{\mathbf{b}} + \ \overrightarrow{\mathbf{c}} ight) \bullet \overrightarrow{\mathbf{q}} + (\overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{a}}) \bullet \overrightarrow{\mathbf{r}}\$ is equal to

Question

Let $\overrightarrow{\mathbf{a}},\ \overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}$ be three non-coplanar vectors, and let $\overrightarrow{\mathbf{p}},\ \overrightarrow{\mathbf{q}}\ and\ \overrightarrow{\mathbf{r}}$ be vectors defined by the relations $\overrightarrow{\mathbf{p}} = \frac{\overrightarrow{\mathbf{b}} 	imes \overrightarrow{\mathbf{c}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack}\ ,\ \overrightarrow{\mathbf{q}} = \frac{\overrightarrow{\mathbf{c}} 	imes \overrightarrow{\mathbf{a}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack}and\ \ \overrightarrow{\mathbf{r}} = \frac{\overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack}\ \$ Then, the value of the expression $\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} ight) \bullet \overrightarrow{\mathbf{p}} + \left( \overrightarrow{\mathbf{b}} + \ \overrightarrow{\mathbf{c}} ight) \bullet \overrightarrow{\mathbf{q}} + (\overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{a}}) \bullet \overrightarrow{\mathbf{r}}\$ is equal to

Dr. Abubacker · Accepted Answer

(d)
$Given,\ \ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{p}} = \frac{\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} 	imes \overrightarrow{\mathbf{c}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack} = 1$
$and\ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{q}} = \overrightarrow{\mathbf{a}}\frac{\overrightarrow{\mathbf{c}} 	imes \overrightarrow{\mathbf{a}}}{\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} ightbrack} = 0$
Similarly, $\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{q}} = \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{r}} = 1,$
$and\ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{r}} = \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{p}} = \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{q}} = \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{p}} = \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{r}} = 0$
$	herefore\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} ight) \bullet \overrightarrow{\mathbf{p}}\left( \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} ight) \bullet \overrightarrow{\mathbf{q}} + (\overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{a}}) \bullet \overrightarrow{\mathbf{r}}$
$= \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{p}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{p}} + \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{q}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{q}} + \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{r}} + \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{r}}$
$= 1 + 1 + 1 = 3$

Explanation

Reviewed by Dr. Abubacker

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