If e1→,e2→ are two unit vectors such that |e1→+e2→|=3, then (2e1→−5e2→). (3e1→+e2→)=

e¯1=e¯2=1Now e→1+e→2=3e→12+e→22+2e¯1⋅e¯2=31+1+2e→1⋅e→2=3e→1⋅e→2=1/2Now (2e¯1−5e¯2).(3e¯1+e¯2)=6e¯12−13e¯1⋅e¯2−5e¯22=6−13(1/2)−5=−112

If e1→,e2→ are two unit vectors such that |e1→+e2→|=3, then (2e1→−5e2→). (3e1→+e2→)=

e¯1=e¯2=1Now e→1+e→2=3e→12+e→22+2e¯1⋅e¯2=31+1+2e→1⋅e→2=3e→1⋅e→2=1/2Now (2e¯1−5e¯2).(3e¯1+e¯2)=6e¯12−13e¯1⋅e¯2−5e¯22=6−13(1/2)−5=−112

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If $\vec{e_{1}}, \vec{e_{2}}$ are two unit vectors such that $| \vec{e_{1}} + \vec{e_{2}} | = \sqrt{3},$ then $(2 \vec{e_{1}} - 5 \vec{e_{2}})$ . $(3 \vec{e_{1}} + \vec{e_{2}}) =$

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$\frac{11}{2}$

$- \frac{11}{2}$

$\frac{9}{2}$

$- \frac{9}{2}$

answer is B.

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Detailed Solution

$|{\bar{e}}_{1}| = |{\bar{e}}_{2}| = 1$
Now $|{\vec{e}}_{1} + {\vec{e}}_{2}| = \sqrt{3}$
$\begin{array}{l} {|{\vec{e}}_{1}|}^{2} + {|{\vec{e}}_{2}|}^{2} + 2 {\bar{e}}_{1} \cdot {\bar{e}}_{2} = 3 \\ 1 + 1 + 2 {\vec{e}}_{1} \cdot {\vec{e}}_{2} = 3 \\ {\vec{e}}_{1} \cdot {\vec{e}}_{2} = 1 / 2 \end{array}$
Now $(2 {\bar{e}}_{1} - 5 {\bar{e}}_{2}) . (3 {\bar{e}}_{1} + {\bar{e}}_{2})$
$\begin{array}{l} = 6 {|{\bar{e}}_{1}|}^{2} - {13 |{\bar{e}}_{1} \cdot {\bar{e}}_{2} - 5| {\bar{e}}_{2}|}^{2} \\ = 6 - 13 (1 / 2) - 5 \\ = - \frac{11}{2} \end{array}$