CBSE 2022 · Region 3 · Set 2 · Q7 · 3 marks
If $\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are three vectors such that $\displaystyle \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$ and $\displaystyle \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=$ $\displaystyle \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{a}} \neq 0$, then show that $\displaystyle \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}$.
OR If $\displaystyle |\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{~b}}|=5,|\overrightarrow{\mathrm{c}}|=4$ and $\displaystyle \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$, then find the value of $\displaystyle (\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})$.
Marking-scheme solution
$\displaystyle \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$
$\displaystyle \Rightarrow \vec{a} \cdot (\vec{b} - \vec{c}) = 0$
$\displaystyle \Rightarrow \vec{a} = 0,\ \vec{b} = \vec{c}$ or $\displaystyle \vec{a} \perp (\vec{b} - \vec{c})$
but $\displaystyle a \neq 0 \Rightarrow \vec{b} = \vec{c}$ or $\displaystyle \vec{a} \perp (\vec{b} - \vec{c})$
$\displaystyle \vec{a} \times \vec{b} = \vec{a} \times \vec{c}$
$\displaystyle \vec{a} \times (\vec{b} - \vec{c}) = 0$
$\displaystyle \vec{a} = 0,\ \vec{b} = \vec{c}$ or $\displaystyle \vec{a} \,||\, (\vec{b} - \vec{c})$
From ($\displaystyle 1$) and
$\displaystyle \vec{a} \neq 0$ given and $\displaystyle \vec{a}$ cannot be both $\displaystyle \perp$ and $\displaystyle \|$ to $\displaystyle (\vec{b} - \vec{c})$
$\displaystyle \therefore\ \vec{b} = \vec{c}$
$\displaystyle \vec{a} + \vec{b} + \vec{c} = 0$ gives $\displaystyle |\vec{a} + \vec{b} + \vec{c}|^2 = 0$
$\displaystyle \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}| + 2\vec{a} \cdot \vec{b} + 2\vec{b} \cdot \vec{c} + 2\vec{c} \cdot \vec{a} = 0$
$\displaystyle \Rightarrow 9 + 25 + 16 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$\displaystyle \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -25$
Vector AlgebraProduct of Two VectorsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.