Let ΔPQR be a triangle. Let →a=→QR,→b=→RP. If |→a|=12,|→b|=4√3 and →b.→c=24, then which of the following is (are) true?
A
|→c|22−|→a|=12
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B
|→c|22+|→a|=30
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C
|→a×→b+→c×→a|=48√3
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D
→a.→b=−72
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Solution
The correct options are A|→c|22−|→a|=12 B|→a×→b+→c×→a|=48√3 D→a.→b=−72 Using vectors summation method, −→a=→b+→c |→a|2=|→c|2+|→b|2−2|→c||→b|cosP |→a|2=|→c|2+|→b|2+2|→c||→b|cos(π−P) |→a|2=|→c|2+|→b|2+2→c⋅→b |→a|2=122−(4√3)2−2×24 →c=4√3 ∴|→c|22−|→a|=12 ............... Option A. −→c=→a+→b |→c|2=|→a|2+|→b|2−2|→a||→b|cosR (∵|→a||→b|cosR=−|→a||→b|cos(π−R)=→a⋅→b) ∴|→c|2=122+(4√3)2+2→a⋅→b ⇒→a⋅→b=−72 ................. Option D. ∵→a⋅→b=−72⇒|→a||→b|cos(π−R)=72⇒R=2π3 ∵|→b|=|→c|=4√3 ∴△PQR is a isosceles triangle, with ∠P=∠Q=π6 Now, |→a×→b+→c×→a|=∣∣∣|→a||→b|sin(π−π6)+|→c||→a|sin(π−π6)∣∣∣ ⇒|→a×→b+→c×→a|=48√3 ............. option C.