\displaystyle \textbf{Question 1:}\quad \text{For any three vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\text{ prove that }[\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]=2[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}].\quad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle [\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]
\displaystyle =\{(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}+\overrightarrow{c})\}\cdot(\overrightarrow{c}+\overrightarrow{a})\qquad [\text{By definition}]
\displaystyle =(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}+\overrightarrow{b}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c})\cdot(\overrightarrow{c}+\overrightarrow{a})\qquad [\text{By distributive law}]
\displaystyle =(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}+\overrightarrow{b}\times\overrightarrow{c})\cdot(\overrightarrow{c}+\overrightarrow{a})\qquad [\because \overrightarrow{b}\times\overrightarrow{b}=\overrightarrow{0}]
\displaystyle =(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c}+(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{a}+(\overrightarrow{a}\times\overrightarrow{c})\cdot\overrightarrow{c}+(\overrightarrow{a}\times\overrightarrow{c})\cdot\overrightarrow{a}+(\overrightarrow{b}\times\overrightarrow{c})\cdot\overrightarrow{c}+(\overrightarrow{b}\times\overrightarrow{c})\cdot\overrightarrow{a}\qquad [\text{By distributive law}]
\displaystyle =[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]+[\overrightarrow{a},\overrightarrow{b},\overrightarrow{a}]+[\overrightarrow{a},\overrightarrow{c},\overrightarrow{c}]+[\overrightarrow{a},\overrightarrow{c},\overrightarrow{a}]+[\overrightarrow{b},\overrightarrow{c},\overrightarrow{c}]+[\overrightarrow{b},\overrightarrow{c},\overrightarrow{a}]
\displaystyle =[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]+[\overrightarrow{b},\overrightarrow{c},\overrightarrow{a}]\qquad [\because \text{Scalar triple product when any two vectors are equal is zero}]
\displaystyle =2[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]\qquad [\because [\overrightarrow{b},\overrightarrow{c},\overrightarrow{a}]=[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]]
\displaystyle \text{Hence, }[\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]=2[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}].

\displaystyle \textbf{Question 2:}\quad \text{Show that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}\text{ are coplanar.}\quad [\text{CBSE 2013, 2014, 2016}]
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar}
\displaystyle \Leftrightarrow [\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=0
\displaystyle \Leftrightarrow 2[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=0
\displaystyle \Leftrightarrow [\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]=0
\displaystyle \Leftrightarrow \overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}\text{ are coplanar.}

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