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Question

Mathematics Question on Vector Algebra

Let PQRPQR be a triangle Let a=QR,b=RPandc=PQ\vec{a} = \overrightarrow{QR}, \, \vec{ b} = \overrightarrow{RP} and \vec{c} = \overrightarrow{PQ}. If a=12,b=43\left|\vec{a}\right| = 12, \left|\vec{b}\right| = 4\sqrt{3} and b.c=24\vec{b}. \vec{c} = 24, then which of the following is (are) true ?

A

c22a=12\frac{\left|\vec{c}\right|^{2}}{ 2} - \left|\vec{a}\right| = 12

B

c22a=30\frac{\left|\vec{c}\right|^{2}}{ 2} - \left|\vec{a}\right| = 30

C

a×b+c×a=483\left|\vec{a}\times\vec{b}+\vec{c}\times\vec{a}\right| = 48\sqrt{3}

D

ab=72\vec{a}\cdot\vec{b}= 72

Answer

ab=72\vec{a}\cdot\vec{b}= 72

Explanation

Solution

ab=24\vec{a}\cdot\vec{b}= 24
a+b+c=0\vec{a}+\vec{b}+\vec{c} = 0
a=bc\vec{a} = -\vec{b}-\vec{c}
144=48+c2+2bc144 = 48+\left|\vec{c}\right|^{2} + 2\vec{b}\cdot\vec{c}
96=c2+2(24)96 = \left|\vec{c}\right|^{2} + 2\left(24\right)
c2=48\left|\vec{c}\right|^{2} = 48
(A)c22a=48212=12\left(A\right) \frac{\left|\vec{c}\right|^{2}}{2} -\left|\vec{a}\right| = \frac{48}{2}-12 = 12
So option (A)\left(A\right) is correct
(B)c22+a=482+12=24\left(B\right) \frac{\left|\vec{c}\right|^{2}}{2}+\left|\vec{a}\right| = \frac{48}{2} + 12 = 24
So option (B)\left(B\right) is not correct
(C)a×b+c×a=a×ba×c\left(C\right) | \vec{a} \times\vec{b} + \vec{c}\times \vec{a} | = | \vec{a}\times \vec{b} - \vec{a}\times\vec{c} |
=a×(bc)= \left|\vec{a}\times\left(\vec{b}-\vec{c}\right)\right|
=(bc)×(bc)= \left|\left(-\vec{b}-\vec{c}\right)\times\left(\vec{b}-\vec{c}\right)\right|
=0+b×cc×b+0= \left|0+\vec{b}\times\vec{c}-\vec{c}\times\vec{b}+0\right|
=2b×c= 2 \left|\vec{b}\times\vec{c}\right|
=2(b2c2(bc)2)= 2\left(\sqrt{\left|b\right|^{2}\left|c\right|^{2}-\left(\vec{b}\cdot\vec{c}\right)^{2}}\right)
=248(48)(24)2= 2\sqrt{48\left(48\right)-\left(24\right)^{2}}
=2×24×3=483= 2\times24\times\sqrt{3} = 48\sqrt{3}
So option (C)\left(C\right) is correct
(D)a+b+c=0\left(D\right) \vec{a}+\vec{b}+\vec{c} = 0
a+b=c\vec{a}+\vec{b} = -\vec{c}
144+48+2ab=48144+48+2\vec{a}\cdot\vec{b} = 48
ab=72\vec{a}\cdot\vec{b} = -72
So option (D)\left(D\right) is correct