Question
Download Solution PDFमाना कि ABCD एक समांतर चतुर्भुज है जिसके विकर्ण P पर प्रतिच्छेदित होते हैं और O मूल है। \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}}\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFधारणा:
- एक समांतर चतुर्भुज के विकर्ण एक दूसरे को प्रतिच्छेदित करते हैं।
गणना:
चूंकि, एक समांतर चतुर्भुज के विकर्ण एक दूसरे को प्रतिच्छेदित करते हैं इसलिए P, AC और BD दोनों का मध्य बिंदु है।
\(\Rightarrow \frac{{\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OC}}} }}{2} = \overrightarrow {{\rm{OP}}} \)
\(\therefore \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OC}}} = 2 \times \overrightarrow {{\rm{OP}}} \) …. (1)
अब
\( \Rightarrow \frac{{\overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OD}}} }}{2} = \overrightarrow {{\rm{OP}}} \)
\(\therefore \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OD}}} = 2 \times \overrightarrow {{\rm{OP}}} \) …. (2)
समीकरण 1 और 2 को जोड़ने पर हमें प्राप्त होता है
\(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} = 4{\rm{\;}}\overrightarrow {{\rm{OP}}} {\rm{\;}}\)
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