Question

Let \[\vec{a} , \vec{b} , \vec{c}\] be three unit vectors, such that \[\left| \vec{a} + \vec{b} + \vec{c} \right|\] =1 and \[\vec{a}\] is perpendicular to \[\vec{b}\]  If \[\vec{c}\] makes angles α and β with \[\vec{a} and \vec{b}\] respectively, then cos α + cos β =

विकल्प

• (a) \[- \frac{3}{2}\]

• (b) \[\frac{3}{2}\]

• (c) 1 

• (d) −1

Answer 1

(d) −1 

\[\text{ Given that } \vec{a} , \vec{b} \text{ and } \vec{c} are \text{ unit } vectors.\]
\[\text{ So },\left| \vec{a} \right|=1,\left| \vec{b} \right|=1and\left| \vec{c} \right|=1.\]
\[\text{ Since } \vec{a} \text{ and } \vec{b} \text{ are mutually perpendicular },\]
\[ \vec{a} . \vec{b} = 0\]
\[\text{ Now },\]
\[\left| \vec{a} + \vec{b} + \vec{c} \right| = 1\]
\[ \Rightarrow \left| \vec{a} + \vec{b} + \vec{c} \right|^2 = 1\]
\[ \Rightarrow \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2 + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 1\]
\[ \Rightarrow 1 + 1 + 1 + 2\left( 0 \right) + 2 \left| \vec{a} \right| \left| \vec{b} \right| \cos \beta + 2 \left| \vec{c} \right| \left| \vec{a} \right| \cos \alpha = 1\]
\[ \Rightarrow 3 + 2 \left( \cos \alpha + \cos \beta \right) = 1\]
\[ \Rightarrow 2 \left( \cos \alpha + \cos \beta \right) = - 2\]
\[ \Rightarrow \cos \alpha + \cos \beta = - 1\]
\[\]