Solution:
Given that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=1$ and $\vec{\text{c}}=1.$
Since $\vec{\text{a}}$ and $\vec{\text{b}}$ are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=0$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}=1$
$\Rightarrow1+1+1+2(0)+2|\vec{\text{a}}|\big|\vec{\text{b}\big|}\cos\beta+2|\vec{\text{c}}||\vec{\text{a}}|\cos\alpha=1$
$\Rightarrow3+2(\cos\alpha+\cos\beta)=1$
$\Rightarrow2(\cos\alpha+\cos\beta)=-2$
$\Rightarrow\cos\alpha+\cos\beta=-1$
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$(1)$ $a+b=3$
$(2)$ $\operatorname{det}\left(\operatorname{adj} M ^2\right)=81$
$(3)$ $(\operatorname{adj} M)^{-1}+\operatorname{adj} M^{-1}=-M$
$(4)$ If $M \left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$, then $\alpha-\beta+\gamma=3$