Use the triangle vector relation
In triangle P Q R PQR P QR
a ⃗ = Q R → , b ⃗ = R P → , c ⃗ = P Q → . \vec a=\overrightarrow{QR},\quad \vec b=\overrightarrow{RP},\quad \vec c=\overrightarrow{PQ}. a = QR , b = R P , c = P Q . Traversing the triangle gives
P Q → + Q R → + R P → = c ⃗ + a ⃗ + b ⃗ = 0 ⃗ . \overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RP}=\vec c+\vec a+\vec b=\vec 0. P Q + QR + R P = c + a + b = 0 . So,
a ⃗ + b ⃗ + c ⃗ = 0 ⇒ c ⃗ = − ( a ⃗ + b ⃗ ) . \vec a+\vec b+\vec c=0 \quad \Rightarrow \quad \vec c=-(\vec a+\vec b). a + b + c = 0 ⇒ c = − ( a + b ) .
Use the given dot product to find a ⃗ ⋅ b ⃗ \vec a\cdot \vec b a ⋅ b
Given:
∣ a ⃗ ∣ = 12 , ∣ b ⃗ ∣ = 4 3 , b ⃗ ⋅ c ⃗ = 24. |\vec a|=12,\qquad |\vec b|=4\sqrt3,\qquad \vec b\cdot \vec c=24. ∣ a ∣ = 12 , ∣ b ∣ = 4 3 , b ⋅ c = 24. Now,
b ⃗ ⋅ c ⃗ = b ⃗ ⋅ ( − ( a ⃗ + b ⃗ ) ) = − a ⃗ ⋅ b ⃗ − ∣ b ⃗ ∣ 2 . \vec b\cdot \vec c=\vec b\cdot (-(\vec a+\vec b))=-\vec a\cdot \vec b-|\vec b|^2. b ⋅ c = b ⋅ ( − ( a + b )) = − a ⋅ b − ∣ b ∣ 2 . Since
∣ b ⃗ ∣ 2 = ( 4 3 ) 2 = 48 , |\vec b|^2=(4\sqrt3)^2=48, ∣ b ∣ 2 = ( 4 3 ) 2 = 48 , we get
24 = − a ⃗ ⋅ b ⃗ − 48. 24=-\vec a\cdot \vec b-48. 24 = − a ⋅ b − 48. Hence,
− a ⃗ ⋅ b ⃗ = 72 ⇒ a ⃗ ⋅ b ⃗ = − 72. -\vec a\cdot \vec b=72 \quad \Rightarrow \quad \vec a\cdot \vec b=-72. − a ⋅ b = 72 ⇒ a ⋅ b = − 72. So Option D is true .
Find ∣ c ⃗ ∣ 2 |\vec c|^2 ∣ c ∣ 2
Using
c ⃗ = − ( a ⃗ + b ⃗ ) , \vec c=-(\vec a+\vec b), c = − ( a + b ) , we have
∣ c ⃗ ∣ 2 = ∣ a ⃗ + b ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 + ∣ b ⃗ ∣ 2 + 2 a ⃗ ⋅ b ⃗ . |\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot \vec b. ∣ c ∣ 2 = ∣ a + b ∣ 2 = ∣ a ∣ 2 + ∣ b ∣ 2 + 2 a ⋅ b . Substitute values:
∣ c ⃗ ∣ 2 = 12 2 + ( 4 3 ) 2 + 2 ( − 72 ) = 144 + 48 − 144 = 48. |\vec c|^2=12^2+(4\sqrt3)^2+2(-72)=144+48-144=48. ∣ c ∣ 2 = 1 2 2 + ( 4 3 ) 2 + 2 ( − 72 ) = 144 + 48 − 144 = 48. Thus,
∣ c ⃗ ∣ 2 2 = 48 2 = 24. \frac{|\vec c|^2}{2}=\frac{48}{2}=24. 2 ∣ c ∣ 2 = 2 48 = 24. Now check options A and B:
∣ c ⃗ ∣ 2 2 − ∣ a ⃗ ∣ = 24 − 12 = 12 \frac{|\vec c|^2}{2}-|\vec a|=24-12=12 2 ∣ c ∣ 2 − ∣ a ∣ = 24 − 12 = 12 True.
∣ c ⃗ ∣ 2 2 + ∣ a ⃗ ∣ = 24 + 12 = 36 ≠ 30 \frac{|\vec c|^2}{2}+|\vec a|=24+12=36\neq 30 2 ∣ c ∣ 2 + ∣ a ∣ = 24 + 12 = 36 = 30 False.
So Option A is true , Option B is false .
Evaluate Option C
Given expression:
∣ a ⃗ × b ⃗ + c ⃗ × a ⃗ ∣ . \left|\vec a\times \vec b+\vec c\times \vec a\right|. a × b + c × a . Use c ⃗ = − ( a ⃗ + b ⃗ ) \vec c=-(\vec a+\vec b) c = − ( a + b )
c ⃗ × a ⃗ = − ( a ⃗ + b ⃗ ) × a ⃗ = − a ⃗ × a ⃗ − b ⃗ × a ⃗ . \vec c\times \vec a=-(\vec a+\vec b)\times \vec a=-\vec a\times \vec a-\vec b\times \vec a. c × a = − ( a + b ) × a = − a × a − b × a . Since a ⃗ × a ⃗ = 0 \vec a\times \vec a=0 a × a = 0 b ⃗ × a ⃗ = − a ⃗ × b ⃗ \vec b\times \vec a=-\vec a\times \vec b b × a = − a × b
c ⃗ × a ⃗ = 0 + a ⃗ × b ⃗ = a ⃗ × b ⃗ . \vec c\times \vec a=0+\vec a\times \vec b=\vec a\times \vec b. c × a = 0 + a × b = a × b . Therefore,
a ⃗ × b ⃗ + c ⃗ × a ⃗ = 2 ( a ⃗ × b ⃗ ) . \vec a\times \vec b+\vec c\times \vec a=2(\vec a\times \vec b). a × b + c × a = 2 ( a × b ) . Hence,
∣ a ⃗ × b ⃗ + c ⃗ × a ⃗ ∣ = 2 ∣ a ⃗ × b ⃗ ∣ . \left|\vec a\times \vec b+\vec c\times \vec a\right|=2|\vec a\times \vec b|. a × b + c × a = 2∣ a × b ∣. Now,
∣ a ⃗ × b ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 ∣ b ⃗ ∣ 2 − ( a ⃗ ⋅ b ⃗ ) 2 . |\vec a\times \vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot \vec b)^2. ∣ a × b ∣ 2 = ∣ a ∣ 2 ∣ b ∣ 2 − ( a ⋅ b ) 2 . So,
∣ a ⃗ × b ⃗ ∣ 2 = 144 ⋅ 48 − ( − 72 ) 2 = 6912 − 5184 = 1728. |\vec a\times \vec b|^2=144\cdot 48-(-72)^2=6912-5184=1728. ∣ a × b ∣ 2 = 144 ⋅ 48 − ( − 72 ) 2 = 6912 − 5184 = 1728. Thus,
∣ a ⃗ × b ⃗ ∣ = 1728 = 24 3 . |\vec a\times \vec b|=\sqrt{1728}=24\sqrt3. ∣ a × b ∣ = 1728 = 24 3 . Therefore,
∣ a ⃗ × b ⃗ + c ⃗ × a ⃗ ∣ = 2 ⋅ 24 3 = 48 3 . \left|\vec a\times \vec b+\vec c\times \vec a\right|=2\cdot 24\sqrt3=48\sqrt3. a × b + c × a = 2 ⋅ 24 3 = 48 3 . So Option C is true .
Final conclusion
The true statements are:
A , C , D \boxed{A,\ C,\ D} A , C , D
Comparison with stored correct answer
Stored correct answer: A , C , D A, C, D A , C , D
My derived answer matches exactly.