To solve this, let's start with the given vector equation:
c ⃗ = a ⃗ − b ⃗ \vec{c} = \vec{a} - \vec{b} c = a − b
Given vectors are:
a → = α i ^ + 4 j ^ + 2 k ^ \overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k} a = α i ^ + 4 j ^ + 2 k ^
b → = 5 i ^ + 3 j ^ + 4 k ^ \overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} b = 5 i ^ + 3 j ^ + 4 k ^
Then, the vector c → \overrightarrow{\mathrm{c}} c is:
c → = ( α i ^ + 4 j ^ + 2 k ^ ) − ( 5 i ^ + 3 j ^ + 4 k ^ ) \overrightarrow{\mathrm{c}} = (\alpha \hat{i} + 4 \hat{j} + 2 \hat{k}) - (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) c = ( α i ^ + 4 j ^ + 2 k ^ ) − ( 5 i ^ + 3 j ^ + 4 k ^ )
c → = ( α − 5 ) i ^ + ( 4 − 3 ) j ^ + ( 2 − 4 ) k ^ \overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + (4 - 3) \hat{j} + (2 - 4) \hat{k} c = ( α − 5 ) i ^ + ( 4 − 3 ) j ^ + ( 2 − 4 ) k ^
c → = ( α − 5 ) i ^ + 1 j ^ − 2 k ^ \overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k} c = ( α − 5 ) i ^ + 1 j ^ − 2 k ^
The area of the triangle formed by vectors a ⃗ \vec{a} a and b ⃗ \vec{b} b is given by the magnitude of the cross product of a ⃗ \vec{a} a and b ⃗ \vec{b} b , divided by 2:
Area = 1 2 ∣ a ⃗ × b ⃗ ∣ = 5 6 \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 5 \sqrt{6} Area = 2 1 ∣ a × b ∣ = 5 6
This implies:
∣ a ⃗ × b ⃗ ∣ = 10 6 |\vec{a} \times \vec{b}| = 10 \sqrt{6} ∣ a × b ∣ = 10 6
Let's find a ⃗ × b ⃗ \vec{a} \times \vec{b} a × b :
a ⃗ × b ⃗ = ∣ i ^ a m p ; j ^ a m p ; k ^ α a m p ; 4 a m p ; 2 5 a m p ; 3 a m p ; 4 ∣ = i ^ ( 4 ⋅ 4 − 2 ⋅ 3 ) − j ^ ( α ⋅ 4 − 2 ⋅ 5 ) + k ^ ( α ⋅ 3 − 4 ⋅ 5 ) \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = \hat{i}(4 \cdot 4 - 2 \cdot 3) - \hat{j}(\alpha \cdot 4 - 2 \cdot 5) + \hat{k}(\alpha \cdot 3 - 4 \cdot 5) a × b = i ^ α 5 am p ; j ^ am p ; 4 am p ; 3 am p ; k ^ am p ; 2 am p ; 4 = i ^ ( 4 ⋅ 4 − 2 ⋅ 3 ) − j ^ ( α ⋅ 4 − 2 ⋅ 5 ) + k ^ ( α ⋅ 3 − 4 ⋅ 5 )
a ⃗ × b ⃗ = i ^ ( 16 − 6 ) − j ^ ( 4 α − 10 ) + k ^ ( 3 α − 20 ) \vec{a} \times \vec{b} = \hat{i}(16 - 6) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20) a × b = i ^ ( 16 − 6 ) − j ^ ( 4 α − 10 ) + k ^ ( 3 α − 20 )
a ⃗ × b ⃗ = i ^ ( 10 ) − j ^ ( 4 α − 10 ) + k ^ ( 3 α − 20 ) \vec{a} \times \vec{b} = \hat{i}(10) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20) a × b = i ^ ( 10 ) − j ^ ( 4 α − 10 ) + k ^ ( 3 α − 20 )
Magnitude of a ⃗ × b ⃗ \vec{a} \times \vec{b} a × b :
∣ a ⃗ × b ⃗ ∣ = 10 2 + ( 4 α − 10 ) 2 + ( 3 α − 20 ) 2 |\vec{a} \times \vec{b}| = \sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2} ∣ a × b ∣ = 1 0 2 + ( 4 α − 10 ) 2 + ( 3 α − 20 ) 2
We know:
10 2 + ( 4 α − 10 ) 2 + ( 3 α − 20 ) 2 = 10 6 \sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2} = 10 \sqrt{6} 1 0 2 + ( 4 α − 10 ) 2 + ( 3 α − 20 ) 2 = 10 6
Squaring both sides:
100 + ( 4 α − 10 ) 2 + ( 3 α − 20 ) 2 = 600 100 + (4\alpha - 10)^2 + (3\alpha - 20)^2 = 600 100 + ( 4 α − 10 ) 2 + ( 3 α − 20 ) 2 = 600
100 + 16 α 2 − 80 α + 100 + 9 α 2 − 120 α + 400 = 600 100 + 16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 600 100 + 16 α 2 − 80 α + 100 + 9 α 2 − 120 α + 400 = 600
25 α 2 − 200 α + 600 = 600 25\alpha^2 - 200\alpha + 600 = 600 25 α 2 − 200 α + 600 = 600
25 α 2 − 200 α = 0 25\alpha^2 - 200\alpha = 0 25 α 2 − 200 α = 0
α 2 − 8 α = 0 \alpha^2 - 8\alpha = 0 α 2 − 8 α = 0
α ( α − 8 ) = 0 \alpha(\alpha - 8) = 0 α ( α − 8 ) = 0
Since α \alpha α is a positive real number:
α = 8 \alpha = 8 α = 8
Then, the vector c ⃗ \vec{c} c is:
c ⃗ = ( 8 − 5 ) i ^ + 1 j ^ − 2 k ^ \vec{c} = (8 - 5)\hat{i} + 1\hat{j} - 2\hat{k} c = ( 8 − 5 ) i ^ + 1 j ^ − 2 k ^
c ⃗ = 3 i ^ + 1 j ^ − 2 k ^ \vec{c} = 3\hat{i} + 1\hat{j} - 2\hat{k} c = 3 i ^ + 1 j ^ − 2 k ^
Magnitude squared of c ⃗ \vec{c} c is:
∣ c ⃗ ∣ 2 = 3 2 + 1 2 + ( − 2 ) 2 |\vec{c}|^2 = 3^2 + 1^2 + (-2)^2 ∣ c ∣ 2 = 3 2 + 1 2 + ( − 2 ) 2
∣ c ⃗ ∣ 2 = 9 + 1 + 4 |\vec{c}|^2 = 9 + 1 + 4 ∣ c ∣ 2 = 9 + 1 + 4
∣ c ⃗ ∣ 2 = 14 |\vec{c}|^2 = 14 ∣ c ∣ 2 = 14
Therefore, the correct option is:
Option A: 14