Use the triangle vector relation
For triangle A B C ABC A B C
A B → + B C → = A C → \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC} A B + B C = A C Equivalently,
A B → = C B → + C A → \overrightarrow{AB}=\overrightarrow{CB}+\overrightarrow{CA} A B = C B + C A since B C → = − C B → \overrightarrow{BC}=-\overrightarrow{CB} B C = − C B A C → = − C A → \overrightarrow{AC}=-\overrightarrow{CA} A C = − C A
Given:
A B → = − 2 i ^ + j ^ + 3 k ^ , C A → = 4 i ^ + 3 j ^ + δ k ^ \overrightarrow{AB}=-2\hat i+\hat j+3\hat k,
\qquad \overrightarrow{CA}=4\hat i+3\hat j+\delta \hat k A B = − 2 i ^ + j ^ + 3 k ^ , C A = 4 i ^ + 3 j ^ + δ k ^ So,
C B → = A B → − C A → \overrightarrow{CB}=\overrightarrow{AB}-\overrightarrow{CA} C B = A B − C A Hence,
C B → = ( − 2 − 4 ) i ^ + ( 1 − 3 ) j ^ + ( 3 − δ ) k ^ \overrightarrow{CB}=(-2-4)\hat i+(1-3)\hat j+(3-\delta)\hat k C B = ( − 2 − 4 ) i ^ + ( 1 − 3 ) j ^ + ( 3 − δ ) k ^ C B → = − 6 i ^ − 2 j ^ + ( 3 − δ ) k ^ \boxed{\overrightarrow{CB}=-6\hat i-2\hat j+(3-\delta)\hat k} C B = − 6 i ^ − 2 j ^ + ( 3 − δ ) k ^ Thus,
α = − 6 , β = − 2 , γ = 3 − δ \alpha=-6,\quad \beta=-2,\quad \gamma=3-\delta α = − 6 , β = − 2 , γ = 3 − δ
Use area of triangle
Area of triangle formed by vectors C A → \overrightarrow{CA} C A C B → \overrightarrow{CB} C B
1 2 ∣ C A → × C B → ∣ = 5 6 \frac12\left|\overrightarrow{CA}\times \overrightarrow{CB}\right|=5\sqrt6 2 1 C A × C B = 5 6 So,
∣ C A → × C B → ∣ = 10 6 \left|\overrightarrow{CA}\times \overrightarrow{CB}\right|=10\sqrt6 C A × C B = 10 6 Now,
C A → = ( 4 , 3 , δ ) , C B → = ( − 6 , − 2 , 3 − δ ) \overrightarrow{CA}=(4,3,\delta),\qquad \overrightarrow{CB}=(-6,-2,3-\delta) C A = ( 4 , 3 , δ ) , C B = ( − 6 , − 2 , 3 − δ ) Compute cross product:
C A → × C B → = ∣ i ^ j ^ k ^ 4 3 δ − 6 − 2 3 − δ ∣ \overrightarrow{CA}\times\overrightarrow{CB}=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
4&3&\delta\\
-6&-2&3-\delta
\end{vmatrix} C A × C B = i ^ 4 − 6 j ^ 3 − 2 k ^ δ 3 − δ = i ^ ( 3 ( 3 − δ ) − δ ( − 2 ) ) − j ^ ( 4 ( 3 − δ ) − δ ( − 6 ) ) + k ^ ( 4 ( − 2 ) − 3 ( − 6 ) ) =\hat i\big(3(3-\delta)-\delta(-2)\big)
-\hat j\big(4(3-\delta)-\delta(-6)\big)
+\hat k\big(4(-2)-3(-6)\big) = i ^ ( 3 ( 3 − δ ) − δ ( − 2 ) ) − j ^ ( 4 ( 3 − δ ) − δ ( − 6 ) ) + k ^ ( 4 ( − 2 ) − 3 ( − 6 ) ) = i ^ ( 9 − 3 δ + 2 δ ) − j ^ ( 12 − 4 δ + 6 δ ) + k ^ ( − 8 + 18 ) =\hat i(9-3\delta+2\delta)-\hat j(12-4\delta+6\delta)+\hat k(-8+18) = i ^ ( 9 − 3 δ + 2 δ ) − j ^ ( 12 − 4 δ + 6 δ ) + k ^ ( − 8 + 18 ) = ( 9 − δ ) i ^ − ( 12 + 2 δ ) j ^ + 10 k ^ =(9-\delta)\hat i-(12+2\delta)\hat j+10\hat k = ( 9 − δ ) i ^ − ( 12 + 2 δ ) j ^ + 10 k ^ So,
∣ C A → × C B → ∣ 2 = ( 9 − δ ) 2 + ( 12 + 2 δ ) 2 + 10 2 \left|\overrightarrow{CA}\times\overrightarrow{CB}\right|^2=(9-\delta)^2+(12+2\delta)^2+10^2 C A × C B 2 = ( 9 − δ ) 2 + ( 12 + 2 δ ) 2 + 1 0 2 This equals
( 10 6 ) 2 = 600 (10\sqrt6)^2=600 ( 10 6 ) 2 = 600 Therefore,
( 9 − δ ) 2 + ( 12 + 2 δ ) 2 + 100 = 600 (9-\delta)^2+(12+2\delta)^2+100=600 ( 9 − δ ) 2 + ( 12 + 2 δ ) 2 + 100 = 600 ( 81 − 18 δ + δ 2 ) + ( 144 + 48 δ + 4 δ 2 ) + 100 = 600 (81-18\delta+\delta^2)+(144+48\delta+4\delta^2)+100=600 ( 81 − 18 δ + δ 2 ) + ( 144 + 48 δ + 4 δ 2 ) + 100 = 600 5 δ 2 + 30 δ + 325 = 600 5\delta^2+30\delta+325=600 5 δ 2 + 30 δ + 325 = 600 5 δ 2 + 30 δ − 275 = 0 5\delta^2+30\delta-275=0 5 δ 2 + 30 δ − 275 = 0 Divide by 5 5 5
δ 2 + 6 δ − 55 = 0 \delta^2+6\delta-55=0 δ 2 + 6 δ − 55 = 0 ( δ + 11 ) ( δ − 5 ) = 0 (\delta+11)(\delta-5)=0 ( δ + 11 ) ( δ − 5 ) = 0 So,
δ = − 11 or 5 \delta=-11\quad \text{or} \quad 5 δ = − 11 or 5 Given δ > 0 \delta>0 δ > 0
δ = 5 \boxed{\delta=5} δ = 5
Find C B → ⋅ C A → \overrightarrow{CB}\cdot\overrightarrow{CA} C B ⋅ C A
Now,
C A → = ( 4 , 3 , 5 ) , C B → = ( − 6 , − 2 , 3 − 5 ) = ( − 6 , − 2 , − 2 ) \overrightarrow{CA}=(4,3,5),\qquad \overrightarrow{CB}=(-6,-2,3-5)=(-6,-2,-2) C A = ( 4 , 3 , 5 ) , C B = ( − 6 , − 2 , 3 − 5 ) = ( − 6 , − 2 , − 2 ) Thus,
C B → ⋅ C A → = ( − 6 ) ( 4 ) + ( − 2 ) ( 3 ) + ( − 2 ) ( 5 ) \overrightarrow{CB}\cdot\overrightarrow{CA}=(-6)(4)+(-2)(3)+(-2)(5) C B ⋅ C A = ( − 6 ) ( 4 ) + ( − 2 ) ( 3 ) + ( − 2 ) ( 5 ) = − 24 − 6 − 10 = − 40 =-24-6-10=-40 = − 24 − 6 − 10 = − 40 So,
C B → ⋅ C A → = − 40 \boxed{\overrightarrow{CB}\cdot\overrightarrow{CA}=-40} C B ⋅ C A = − 40
Compare with options
The computed value is − 40 -40 − 40 not present in the options 60 , 54 , 120 , 108 60, 54, 120, 108 60 , 54 , 120 , 108
Check if the intended quantity might be ( C B → ⋅ C A → ) 2 \left(\overrightarrow{CB}\cdot\overrightarrow{CA}\right)^2 ( C B ⋅ C A ) 2
( − 40 ) 2 = 1600 (-40)^2=1600 ( − 40 ) 2 = 1600 Not an option.
Alternative check using identity
Using
∣ C A → × C B → ∣ 2 = ∣ C A ∣ 2 ∣ C B ∣ 2 − ( C A ⋅ C B ) 2 \left|\overrightarrow{CA}\times\overrightarrow{CB}\right|^2
=|CA|^2|CB|^2-(CA\cdot CB)^2 C A × C B 2 = ∣ C A ∣ 2 ∣ C B ∣ 2 − ( C A ⋅ C B ) 2 we have
∣ C A ∣ 2 = 4 2 + 3 2 + 5 2 = 50 , |CA|^2=4^2+3^2+5^2=50, ∣ C A ∣ 2 = 4 2 + 3 2 + 5 2 = 50 , ∣ C B ∣ 2 = ( − 6 ) 2 + ( − 2 ) 2 + ( − 2 ) 2 = 44 |CB|^2=(-6)^2+(-2)^2+(-2)^2=44 ∣ C B ∣ 2 = ( − 6 ) 2 + ( − 2 ) 2 + ( − 2 ) 2 = 44 Then,
∣ C A ∣ 2 ∣ C B ∣ 2 − ( C A ⋅ C B ) 2 = 50 ⋅ 44 − ( C A ⋅ C B ) 2 |CA|^2|CB|^2-(CA\cdot CB)^2=50\cdot 44-(CA\cdot CB)^2 ∣ C A ∣ 2 ∣ C B ∣ 2 − ( C A ⋅ C B ) 2 = 50 ⋅ 44 − ( C A ⋅ C B ) 2 Since area = 5 6 =5\sqrt6 = 5 6 10 6 10\sqrt6 10 6
600 = 2200 − ( C A ⋅ C B ) 2 600=2200-(CA\cdot CB)^2 600 = 2200 − ( C A ⋅ C B ) 2 ( C A ⋅ C B ) 2 = 1600 (CA\cdot CB)^2=1600 ( C A ⋅ C B ) 2 = 1600 C A ⋅ C B = ± 40 CA\cdot CB=\pm 40 C A ⋅ C B = ± 40 Direct computation gives − 40 -40 − 40
Hence the question/options appear inconsistent.
Final derived answer: − 40 \boxed{-40} − 40