JEE Mains · Maths · STD 12 - 10. vector algebra
In a parallelogram \(ABCD\), \(\left| {\overline {AB} } \right| = a\,,\,\left| {\overline {AD} } \right| = b\) and \(\left| {\overline {AC} } \right| = c\) then \(\overline {DA} \). \(\overline {AB} \) has the value
- A \(\frac{1}{2}\left( {{a^2} + {b^2} + {c^2}} \right)\)
- B \(\frac{1}{2}\left( {{a^2} - {b^2} + {c^2}} \right)\)
- C \(\frac{1}{2}\left( {{a^2} + {b^2} - {c^2}} \right)\)
- D \(\frac{1}{3}\left( {{a^2} + {b^2} - {c^2}} \right)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\left( {{a^2} + {b^2} - {c^2}} \right)\)
Step-by-step Solution
Detailed explanation
Let \(|\overline {{\rm{AB}}} | = {\rm{a}},|\overline {{\rm{AD}}} | = {\rm{band}}|\overline {{\rm{AC}}} | = {\rm{c}}\) We have \(\overline {{\rm{AB}}} + \overline {{\rm{AD}}} = \overline {{\rm{AC}}} \) On squaring both the side, we get…
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