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If in a right-angled triangle $\mathrm{ABC}$, hypotenuse $\mathrm{AC}=\mathrm{p}$, then what is $\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}$ equal to ?
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The correct answer is:
$\mathrm{p}^{2}$
Hypotenuse, $\mathrm{AC}=\mathrm{P}$ $\mathrm{BC}$ is perpendicular to $\mathrm{AB}$.
$\therefore \overrightarrow{B C} \cdot \overrightarrow{B A}=0$
$\therefore \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}$
$=\overrightarrow{A B} \cdot \overrightarrow{A C}+0+\overrightarrow{A C} \cdot \overrightarrow{B C}=\overrightarrow{A C}(\overrightarrow{A B}+\overrightarrow{B C})$
$=\overrightarrow{A C} \cdot \overrightarrow{A C}=\overrightarrow{A C}^{2}=\mathrm{P}^{2}$
$\therefore \overrightarrow{B C} \cdot \overrightarrow{B A}=0$
$\therefore \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}$
$=\overrightarrow{A B} \cdot \overrightarrow{A C}+0+\overrightarrow{A C} \cdot \overrightarrow{B C}=\overrightarrow{A C}(\overrightarrow{A B}+\overrightarrow{B C})$
$=\overrightarrow{A C} \cdot \overrightarrow{A C}=\overrightarrow{A C}^{2}=\mathrm{P}^{2}$
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