\displaystyle \textbf{Question 1: } \text{Find }(\overrightarrow{a}+3\overrightarrow{b})\cdot(2\overrightarrow{a}-\overrightarrow{b}), \text{ if }\overrightarrow{a}=\widehat{i}+\widehat{j}+2\widehat{k} \text{ and }\overrightarrow{b}=3\widehat{i}+2\widehat{j}-\widehat{k}.  \hspace{12.0cm}[\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\overrightarrow{a}=\widehat{i}+\widehat{j}+2\widehat{k}\ \text{and }\overrightarrow{b}=3\widehat{i}+2\widehat{j}-\widehat{k}
\displaystyle \therefore\ \overrightarrow{a}+3\overrightarrow{b}=(\widehat{i}+\widehat{j}+2\widehat{k})+3(3\widehat{i}+2\widehat{j}-\widehat{k})=10\widehat{i}+7\widehat{j}-\widehat{k}
\displaystyle \text{and, }\ 2\overrightarrow{a}-\overrightarrow{b}=2(\widehat{i}+\widehat{j}+2\widehat{k})-(3\widehat{i}+2\widehat{j}-\widehat{k})=-\widehat{i}+0\widehat{j}+5\widehat{k}
\displaystyle \therefore\ (\overrightarrow{a}+3\overrightarrow{b})\cdot(2\overrightarrow{a}-\overrightarrow{b})=(10\widehat{i}+7\widehat{j}-\widehat{k})\cdot(-\widehat{i}+0\widehat{j}+5\widehat{k})
\displaystyle =(10)(-1)+(7)(0)+(-1)(5)=-10+0-5=-15

\displaystyle \textbf{Question 2: } \text{If } \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\ |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=5,\ |\overrightarrow{c}|=7, \\ \text{ find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}. \hspace{6.0cm}[\text{CBSE 2008, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta\text{ be the angle between vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}.
\displaystyle \text{We have, }
\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow\ \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|-\overrightarrow{c}\right|
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}\right|^{2}=\left|\overrightarrow{c}\right|^{2}
\displaystyle \Rightarrow\ \left|\overrightarrow{a}\right|^{2}+\left|\overrightarrow{b}\right|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}=\left|\overrightarrow{c}\right|^{2}
\displaystyle \Rightarrow\ \left|\overrightarrow{a}\right|^{2}+\left|\overrightarrow{b}\right|^{2}+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos\theta=\left|\overrightarrow{c}\right|^{2}
\displaystyle \Rightarrow\ 9+25+2(3)(5)\cos\theta=49
\displaystyle \Rightarrow\ \cos\theta=\frac{1}{2}\Rightarrow\ \theta=\frac{\pi}{3}

\displaystyle \textbf{Question 3: } \text{If }\overrightarrow{a},\overrightarrow{b}\text{ are two vectors such that }\left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{a}\right|,\ \text{then prove that }2\overrightarrow{a}+\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{b}.\ \text{[CBSE 2013]} 
\displaystyle \text{Answer:}
\displaystyle \text{We have, }
\displaystyle \left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{a}\right|
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}\right|^{2}=\left|\overrightarrow{a}\right|^{2}\ \ \text{[Squaring both sides]}
\displaystyle \Rightarrow\ \left|\overrightarrow{a}\right|^{2}+\left|\overrightarrow{b}\right|^{2}+2(\overrightarrow{a}\cdot\overrightarrow{b})=\left|\overrightarrow{a}\right|^{2}
\displaystyle \Rightarrow\ \left|\overrightarrow{b}\right|^{2}+2(\overrightarrow{a}\cdot\overrightarrow{b})=0\qquad\text{(i)}
\displaystyle \text{Now, }\ (2\overrightarrow{a}+\overrightarrow{b})\cdot\overrightarrow{b}=2(\overrightarrow{a}\cdot\overrightarrow{b})+(\overrightarrow{b}\cdot\overrightarrow{b})=2(\overrightarrow{a}\cdot\overrightarrow{b})+\left|\overrightarrow{b}\right|^{2}=0\ \ \text{[Using (i)]}
\displaystyle \text{Hence, }(2\overrightarrow{a}+\overrightarrow{b})\text{ is perpendicular to }\overrightarrow{b}.

\displaystyle \textbf{Question 4: } \text{The scalar product of the vector } \overrightarrow a=\widehat i+\widehat j+\widehat k \text{ with a unit vector along} \\ \text{the sum of the vectors } \overrightarrow b=2\widehat i+4\widehat j-5\widehat k \text{ and } \overrightarrow c=\lambda\widehat i+2\widehat j+3\widehat k \text{ is equal to }1. \text{ Find the value of } \\ \lambda\text{ and hence find the unit vector along } \overrightarrow b+\overrightarrow c. [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}\text{ and }\overrightarrow{c}=\lambda\widehat{i}+2\widehat{j}+3\widehat{k}.\ \text{Then,}
\displaystyle \overrightarrow{b}+\overrightarrow{c}=(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}
\displaystyle \text{Let }\overrightarrow{r}\text{ denote the unit vector along }\overrightarrow{b}+\overrightarrow{c}.\ \text{Then,}
\displaystyle \overrightarrow{r}=\frac{\overrightarrow{b}+\overrightarrow{c}}{\left|\overrightarrow{b}+\overrightarrow{c}\right|}=\frac{(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{(2+\lambda)^{2}+36+4}}=\frac{(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{(2+\lambda)^{2}+40}}\qquad\text{(i)}
\displaystyle \text{Now, }\ (\widehat{i}+\widehat{j}+\widehat{k})\cdot\overrightarrow{r}=1\qquad\text{[Given]}
\displaystyle \Rightarrow\ \frac{(\widehat{i}+\widehat{j}+\widehat{k})\cdot\{(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}\}}{\sqrt{(2+\lambda)^{2}+40}}=1
\displaystyle \Rightarrow\ (\widehat{i}+\widehat{j}+\widehat{k})\cdot\{(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}\}=\sqrt{(2+\lambda)^{2}+40}
\displaystyle \Rightarrow\ \cos\theta=\frac{(a\widehat{i}+a\widehat{j}+a\widehat{k})\cdot(-a\widehat{i}+a\widehat{j}+a\widehat{k})}{\sqrt{a^{2}+a^{2}+a^{2}}\sqrt{(-a)^{2}+a^{2}+a^{2}}}
\displaystyle \Rightarrow\ \cos\theta=\frac{-a^{2}+a^{2}+a^{2}}{\sqrt{3a}\sqrt{3a}}=\frac{1}{3}
\displaystyle \Rightarrow\ \theta=\cos^{-1}\!\left(\frac{1}{3}\right)
\displaystyle \text{Similarly, angle }\theta\text{ between the other pairs of diagonals is }\cos^{-1}\!\left(\frac{1}{3}\right).

\displaystyle \textbf{Question 5: }\ \ \text{If with reference to a right handed system of mutually perpendicular unit vectors } \\ \widehat{i},\widehat{j},\widehat{k},\ \text{we have }\overrightarrow{a}=3\widehat{i}-\widehat{j},\ \text{and }\overrightarrow{b}=2\widehat{i}+\widehat{j}-3\widehat{k}.\ \text{Express }\overrightarrow{b}\text{ in the form } \\ \overrightarrow{b}=\overrightarrow{b}_{1}+\overrightarrow{b}_{2},\ \text{where }\overrightarrow{b}_{1}\text{ is parallel to }\overrightarrow{a}\text{ and }\overrightarrow{b}_{2}\text{ is perpendicular to }\overrightarrow{a}.\ \text{[CBSE 2012, 2013]}

\displaystyle \text{Answer:}
\displaystyle \text{It is given that }\overrightarrow{b}_{1}\text{ is parallel to }\overrightarrow{a}.\ \text{Therefore,}
\displaystyle \overrightarrow{b}_{1}=\lambda\overrightarrow{a}\ \text{for some scalar }\lambda\qquad\text{(i)}
\displaystyle \text{It is also given that}
\displaystyle \overrightarrow{b}=\overrightarrow{b}_{1}+\overrightarrow{b}_{2}\Rightarrow\overrightarrow{b}_{2}=\overrightarrow{b}-\overrightarrow{b}_{1}=\overrightarrow{b}-\lambda\overrightarrow{a}\qquad\text{(ii)}
\displaystyle \text{It is also given that }\overrightarrow{b}_{2}\text{ is perpendicular to }\overrightarrow{a}.\ \text{Therefore,}
\displaystyle \overrightarrow{b}_{2}\cdot\overrightarrow{a}=0\Rightarrow(\overrightarrow{b}-\lambda\overrightarrow{a})\cdot\overrightarrow{a}=0\Rightarrow\overrightarrow{b}\cdot\overrightarrow{a}-\lambda(\overrightarrow{a}\cdot\overrightarrow{a})=0\Rightarrow\lambda=\frac{\overrightarrow{b}\cdot\overrightarrow{a}}{\overrightarrow{a}\cdot\overrightarrow{a}}\qquad\text{(iii)}
\displaystyle \text{Now, }\overrightarrow{a}=3\widehat{i}-\widehat{j}\ \text{and }\overrightarrow{b}=2\widehat{i}+\widehat{j}-3\widehat{k}
\displaystyle \Rightarrow\ \overrightarrow{b}\cdot\overrightarrow{a}=6-1+0=5\ \text{and }\overrightarrow{a}\cdot\overrightarrow{a}=9+1=10
\displaystyle \text{Substituting these values in (iii), we get}
\displaystyle \lambda=\frac{\overrightarrow{b}\cdot\overrightarrow{a}}{\overrightarrow{a}\cdot\overrightarrow{a}}=\frac{5}{10}=\frac{1}{2}
\displaystyle \therefore\ \overrightarrow{b}_{1}=\lambda\overrightarrow{a}\Rightarrow\overrightarrow{b}_{1}=\frac{1}{2}(3\widehat{i}-\widehat{j})=\frac{3}{2}\widehat{i}-\frac{1}{2}\widehat{j}
\displaystyle \text{and, }\overrightarrow{b}_{2}=\overrightarrow{b}-\overrightarrow{b}_{1}\Rightarrow\overrightarrow{b}_{2}=(2\widehat{i}+\widehat{j}-3\widehat{k})-\left(\frac{3}{2}\widehat{i}-\frac{1}{2}\widehat{j}\right)=\frac{1}{2}\widehat{i}+\frac{3}{2}\widehat{j}-3\widehat{k}
\displaystyle \text{Hence, }\overrightarrow{b}_{1}=\frac{3}{2}\widehat{i}-\frac{1}{2}\widehat{j}\ \text{and }\overrightarrow{b}_{2}=\frac{1}{2}\widehat{i}+\frac{3}{2}\widehat{j}-3\widehat{k}.

\displaystyle \textbf{Question 6: } \text{Find }x\text{ such that the angle between the vectors} \overrightarrow{a}=2x^{2}\widehat{i}+4x\widehat{j}+\widehat{k} \text{ and } \\ \overrightarrow{b}=7\widehat{i}-2\widehat{j}+x\widehat{k} \text{ is obtuse} \hspace{7.0cm}[\text{CBSE 2013}]
\displaystyle \text{Answer:}
\displaystyle \text{The angle }\theta\text{ between vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is given by }\cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}
\displaystyle \text{For the angle }\theta\text{ to be obtuse, we must have}
\displaystyle \cos\theta<0
\displaystyle \Rightarrow\ \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}<0
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot\overrightarrow{b}<0\qquad[\because\ \left|\overrightarrow{a}\right|,\left|\overrightarrow{b}\right|>0]
\displaystyle \Rightarrow\ 14x^{2}-8x+x<0\Rightarrow7x(2x-1)<0\Rightarrow x(2x-1)<0\Rightarrow0<x<\frac{1}{2}
\displaystyle \text{Hence, the angle between the given vectors is obtuse if }x\in(0,\frac{1}{2}).

\displaystyle \textbf{Question 7: } \text{If }\overrightarrow a,\overrightarrow b,\overrightarrow c \text{ are three mutually perpendicular vectors} \\ \text{of equal magnitude, prove that } \overrightarrow a+\overrightarrow b+\overrightarrow c \text{ is equally inclined with } \overrightarrow a,\overrightarrow b,\overrightarrow c. \\ \text{ Also, find the angle. } \hspace{4.0cm}[\text{CBSE 2005, 2011, 2013, 2017}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=\lambda\ \text{(say).}
\displaystyle \text{Since }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are mutually perpendicular vectors. Therefore,}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0\qquad\text{(i)}
\displaystyle \text{We know that }\left|\overrightarrow{x}\right|^{2}=\overrightarrow{x}\cdot\overrightarrow{x}
\displaystyle \therefore\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=\overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}+\overrightarrow{c}\cdot\overrightarrow{c}+2\overrightarrow{a}\cdot\overrightarrow{b}+2\overrightarrow{b}\cdot\overrightarrow{c}+2\overrightarrow{c}\cdot\overrightarrow{a}
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=\left|\overrightarrow{a}\right|^{2}+\left|\overrightarrow{b}\right|^{2}+\left|\overrightarrow{c}\right|^{2}\qquad[\text{Using (i)}]
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=3\lambda^{2}\qquad[\because\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=\lambda]
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\sqrt{3}\lambda\qquad\text{(ii)}
\displaystyle \text{Suppose }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\text{ makes angles }\theta_{1},\theta_{2},\theta_{3}\text{ with }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively. Then,}
\displaystyle \cos\theta_{1}=\frac{\overrightarrow{a}\cdot(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})}{\left|\overrightarrow{a}\right|\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}=\frac{\overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{a}\cdot\overrightarrow{c}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}
\displaystyle \Rightarrow\ \cos\theta_{1}=\frac{\left|\overrightarrow{a}\right|^{2}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}=\frac{\left|\overrightarrow{a}\right|}{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}=\frac{\lambda}{\sqrt{3}\lambda}=\frac{1}{\sqrt{3}}\qquad[\text{Using (ii)}]
\displaystyle \Rightarrow\ \theta_{1}=\cos^{-1}\!\left(\frac{1}{\sqrt{3}}\right)
\displaystyle \text{Similarly, we obtain}
\displaystyle \theta_{2}=\cos^{-1}\!\left(\frac{1}{\sqrt{3}}\right)\ \text{and }\theta_{3}=\cos^{-1}\!\left(\frac{1}{\sqrt{3}}\right)
\displaystyle \therefore\ \theta_{1}=\theta_{2}=\theta_{3}
\displaystyle \text{Hence, }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\text{ is equally inclined with }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}.

\displaystyle \textbf{Question 8: } \text{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ be three vectors of magnitudes }3,4\text{ and }5\text{ respectively. If each one is}
\displaystyle \text{perpendicular to the sum of the other two vectors, prove that }\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=5\sqrt{2}.\ \text{[CBSE 2010, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=4\ \text{and }\left|\overrightarrow{c}\right|=5
\displaystyle \text{It is given that}
\displaystyle \overrightarrow{a}\perp(\overrightarrow{b}+\overrightarrow{c}),\ \overrightarrow{b}\perp(\overrightarrow{c}+\overrightarrow{a})\ \text{and }\overrightarrow{c}\perp(\overrightarrow{a}+\overrightarrow{b})
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot(\overrightarrow{b}+\overrightarrow{c})=0,\ \overrightarrow{b}\cdot(\overrightarrow{c}+\overrightarrow{a})=0\ \text{and }\overrightarrow{c}\cdot(\overrightarrow{a}+\overrightarrow{b})=0
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{a}\cdot\overrightarrow{c}=0,\ \overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{b}\cdot\overrightarrow{a}=0\ \text{and }\overrightarrow{c}\cdot\overrightarrow{a}+\overrightarrow{c}\cdot\overrightarrow{b}=0
\displaystyle \text{Adding all these, we get}
\displaystyle 2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0\Rightarrow\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=0\qquad\text{(i)}
\displaystyle \text{Now, }\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=\left|\overrightarrow{a}\right|^{2}+\left|\overrightarrow{b}\right|^{2}+\left|\overrightarrow{c}\right|^{2}+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=3^{2}+4^{2}+5^{2}+0\qquad[\text{Using (i)}]
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\sqrt{50}=5\sqrt{2}

\displaystyle \textbf{Question 9: } \text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are unit vectors such that }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\ \text{find the value of }\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}.\ \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1\ \text{and }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}
\displaystyle \text{Now, }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=0
\displaystyle \Rightarrow\ \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|^{2}=0
\displaystyle \Rightarrow\ \left|\overrightarrow{a}\right|^{2}+\left|\overrightarrow{b}\right|^{2}+\left|\overrightarrow{c}\right|^{2}+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0
\displaystyle \Rightarrow\ 3+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0\qquad[\because\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1]
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-\frac{3}{2}

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