Let a , b , c be three unit vectors, such that | a + b + c |=1 an… - Vidyadip

Question

Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be three unit vectors, such that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$ and $\vec{\text{a}}$ is perpendicular to $\vec{\text{b}}.$ If $\vec{\text{c}}$ makes angle $\alpha$ and $\beta$ with $\vec{\text{a}}$ and $\vec{\text{b}}$ respectively, then $\cos\alpha+\cos\beta=$

$-\frac{3}{2}$
$\frac{3}{2}$
$1$
$-1$

✓

Answer

$-1$

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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