Let a=α i+2 j- k and b=-2 i+α j+ k, where α ∈ R. If the area of the parallelogram whose adjacent sides are represented by the vectors a and b is…

Given,

$\vec{a}=\alpha \hat{i}+2\hat{j}-\hat{k},\vec{b}=-2\hat{i}+\alpha \hat{j}+\hat{k}$a=αi^+2j^​−k^,b=−2i^+αj^​+k^,

Area of parallelogram $=|\hat{a}\times \hat{b}|$=∣a^×b^∣

$$\hat{a}\times \hat{b}=\sqrt{{(\alpha +2)}^{2}+{(\alpha -2)}^{2}+{({\alpha }^{2}+4)}^{2}}$$a^×b^=(α+2)2+(α−2)2+(α2+4)2​

Given $|\hat{a}\times \hat{b}|=\sqrt{15({\alpha }^{2}+4)}$∣a^×b^∣=15(α2+4)​

So, $2({\alpha }^{2}+4)+{({\alpha }^{2}+4)}^{2}=15({\alpha }^{2}+4)$2(α2+4)+(α2+4)2=15(α2+4)

$$\Rightarrow {({\alpha }^{2}+4)}^{2}=13({\alpha }^{2}+4)$$⇒(α2+4)2=13(α2+4)

$$\Rightarrow {\alpha }^{2}+4=13\therefore {\alpha }^{2}=9$$⇒α2+4=13∴α2=9

Now, $2{|\vec{a}|}^{2}+(\vec{a}\cdot \vec{b}){|\vec{b}|}^{2}$2∣a∣2+(a⋅b)∣b∣2

$${|\vec{a}|}^{2}={\alpha }^{2}+4+1={\alpha }^{2}+5$$∣a∣2=α2+4+1=α2+5

$${|\vec{b}|}^{2}=4+{\alpha }^{2}+1={\alpha }^{2}+5$$∣b∣2=4+α2+1=α2+5

$$\vec{a}\cdot \vec{b}=-2\alpha +2\alpha -1=-1$$a⋅b=−2α+2α−1=−1

$$\therefore 2{|\vec{a}|}^{2}+(\vec{a}\cdot \vec{b}){|\vec{b}|}^{2}$$∴2∣a∣2+(a⋅b)∣b∣2

$$2({\alpha }^{2}+5)-1({\alpha }^{2}+5)={\alpha }^{2}+5=14$$2(α2+5)−1(α2+5)=α2+5=14