यदि एक त्रिभुज ABC के शीर्ष A,B,C क्रमशः (1, 1, 3), (-1, 0, 0),(0, 1, 2) है, तो ∠ABC निर्धारित करें। (∠ABC सदिश  \(\overrightarrow {BA} \) और  \(\overrightarrow {BC} \)के बीच का कोण है)

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AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 1)
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  1. \({{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{{10}}{{102}})\)
  2. \({{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt {14} x\sqrt 6 }})\)
  3. \({{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt {14} x\sqrt 9 }})\)
  4. \({{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt 84}})\)

Answer (Detailed Solution Below)

Option 4 : \({{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt 84}})\)
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दिया गया है:

त्रिभुज ABC के शीर्ष  A,B,C क्रमशः (1, 1, 3), (-1, 0, 0),(0, 1, 2) है।

संकल्पना:

\(\vec{PQ}\) = p.v.(\(\vec{Q}\)) - p.v.(\(\vec{P}\))

सूत्र:

दो सदिश \(\vec{P}\) और \(\vec{Q}\) के बीच का कोण निम्न द्वारा दिया जाता है : 

\(\theta = cos^{-1}(\frac{\vec{P}.\vec{Q}}{PQ})\)

हल:

\(\overrightarrow {BA} \) = (1, 1, 3) - (-1, 0, 0)

= 2î + ĵ + 3k̂ 

\(\overrightarrow {BC} \) = (0, 1, 2) - (-1, 0, 0) = î + ĵ + 2k̂ 

∴ \(\theta = cos^{-1}(\frac{\vec{BA}.\vec{BC}}{BA.BC})\)

\(\Rightarrow \theta = cos^{-1}(\frac{9}{\sqrt 14 \sqrt 6})\))

\(\Rightarrow \theta ={{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt 84}})\)

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