JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 16)
In a triangle PQR, let a = QR, b = RP, and c = PQ. If |a| = 3, |b| = 4
and $${{a\,.(\,c - \,b)} \over {c\,.\,(a - \,b)}} = {{|a|} \over {|a| + |b|}}$$, then the value of |a $$ \times $$ b|2 is ......
and $${{a\,.(\,c - \,b)} \over {c\,.\,(a - \,b)}} = {{|a|} \over {|a| + |b|}}$$, then the value of |a $$ \times $$ b|2 is ......
Answer
108
Explanation
It is given that, in a $$\Delta $$PQR
PQ = c, QR = a and RP = b, so a + b + c = 0
$$ \therefore $$ |a|2 + a . b + a . c = 0 ....(i)
and a . b + |b|2 + c . b = 0 ....(ii)
and, also given that |a| = 3, |b| = 4
and $${{a\,.\,c - a\,.\,b} \over {c\,.\,a - c\,.\,b}} = {{|a|} \over {|a| + |b|}} = {3 \over 7}$$
$$ \Rightarrow 7[(a\,.\,c) - (a\,.\,b)] = 3[(c\,.\,a) - (c\,.\,b)]$$
from Eqs. (i) and (ii), on putting the values of a . c and c . b, we get,
$$ \Rightarrow 7[ - 9 - (a\,.\,b) - (a\,.\,b)] = 3[ - 9 - (a\,.\,b) + 16 + (a\,.\,b)]$$
$$ \Rightarrow 7[ - 9 - 2(a\,.\,b)] = 3 \times 7$$
$$ \Rightarrow a\,.\,b = - 6$$
$$ \because $$ $$|a \times b{|^2} = {(|a||b|)^2} - {(a\,.\,b)^2}$$
= $$144 - 36 = 108$$
PQ = c, QR = a and RP = b, so a + b + c = 0
$$ \therefore $$ |a|2 + a . b + a . c = 0 ....(i)
and a . b + |b|2 + c . b = 0 ....(ii)
and, also given that |a| = 3, |b| = 4
and $${{a\,.\,c - a\,.\,b} \over {c\,.\,a - c\,.\,b}} = {{|a|} \over {|a| + |b|}} = {3 \over 7}$$
$$ \Rightarrow 7[(a\,.\,c) - (a\,.\,b)] = 3[(c\,.\,a) - (c\,.\,b)]$$
from Eqs. (i) and (ii), on putting the values of a . c and c . b, we get,
$$ \Rightarrow 7[ - 9 - (a\,.\,b) - (a\,.\,b)] = 3[ - 9 - (a\,.\,b) + 16 + (a\,.\,b)]$$
$$ \Rightarrow 7[ - 9 - 2(a\,.\,b)] = 3 \times 7$$
$$ \Rightarrow a\,.\,b = - 6$$
$$ \because $$ $$|a \times b{|^2} = {(|a||b|)^2} - {(a\,.\,b)^2}$$
= $$144 - 36 = 108$$
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