CBSE 2024 · Region 3 · Set 1 · Q25 · 2 marks
Let $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ be two non-zero vectors. Prove that $\displaystyle |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \leq|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|$. State the condition under which equality holds, i.e., $\displaystyle |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|$.
Marking-scheme solution
$$|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=|\vec{\mathrm{a}}||\vec{\mathrm{b}}||\sin \theta|As, $\displaystyle 0 \leq|\sin \theta| \leq 1$
$\displaystyle \left.\Rightarrow\right|^{\vec{\mathrm{a}}}| | \vec{\mathrm{b}}| | \sin \theta\left|\leq\left.\right|^{\vec{\mathrm{a}}}\right||\vec{\mathrm{b}}|$
$\displaystyle \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \leq|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|$
For equality, $\displaystyle \sin \theta=1 \Rightarrow \theta=\frac{\pi}{2} \Rightarrow \vec{\mathrm{a}}$ is perpendicular to $\displaystyle \vec{\mathrm{b}}$.
Vector AlgebraProduct of Two VectorsUnderstandvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.