- $0$
- $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
- $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
- $-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
Solution:
$\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big[\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{a}}+\vec{\text{c}}\big)\big]$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=0+0+\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]+\big[\vec{\text{b}}\vec{\text{a}}\vec{\text{c}}\big]+0+0+0+\big[\vec{\text{c}}\vec{\text{b}}\vec{\text{a}}\big]+0$
$=-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
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Let $f(x)=\left\{\begin{array}{cc}\frac{x}{|x|} g(x), & x \neq 0 \\ 0, & x=0\end{array}\right.$
and $h(x)=e^{\text {ld }}$ for all $x \in R$. Let $( f \circ h )(x)$ denote $f(h(x))$ and $( h \circ f )( x )$ denote $h(f(x))$. Then which of the following is (are) true?
$(A)$ $f$ is differentiable at $x=0$
$(B)$ $h$ is differentiable at $x=0$
$(C)$ $f \circ h$ is differentiable at $x=0$
$(D)$ $h \circ f$ is differentiable at $x=0$