Let a be a non-zero vector parallel to the line of intersection of the two planes described by i+ j, i+ k and i- j, j- k. If θ is the angle between…

Given,

$\vec{a}$a be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i}+\hat{j},\hat{i}+\hat{k}$i^+j^​,i^+k^,

So finding the normal vector to the above plane, we get,

$${\vec{n}}_{1}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{matrix}|=\hat{i}-\hat{j}-\hat{k}$$n1​=∣i^11​j^​10​k^01​∣=i^−j^​−k^

And normal vector to planes $\hat{i}-\hat{j},\hat{j}-\hat{k}$i^−j^​,j^​−k^ will be,

$${\vec{n}}_{2}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 0 & 1 & -1\end{matrix}|=\hat{i}+\hat{j}+\hat{k}$$n2​=∣i^10​j^​−11​k^0−1​∣=i^+j^​+k^

Now $\vec{a}=\lambda ({\vec{n}}_{1}\times {\vec{n}}_{2})$a=λ(n1​×n2​)

So, direction ratio of $\vec{a}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 1 & 1 & 1\end{matrix}|$a=∣i^11​j^​−11​k^−11​∣$=-2\hat{j}+2\hat{k}$=−2j^​+2k^

Hence, direction ratio of $\vec{a}=(0,-2,2)=(0,-1,1)$a=(0,−2,2)=(0,−1,1)

Now given, $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$b=2i^−2j^​+k^

And $\vec{a}=\lambda (-\hat{j}+\hat{k})$a=λ(−j^​+k^)

So, $\vec{a}\cdot \vec{b}=6=\lambda (2+1)$a⋅b=6=λ(2+1)

$$\Rightarrow \lambda =2$$⇒λ=2

$$\therefore \vec{a}=-2\hat{j}+2\hat{k}$$∴a=−2j^​+2k^

Now we know that, $\vec{a}\cdot \vec{b}=|\bar{a}||b|\mathrm{cos}\theta$a⋅b=∣aˉ∣∣b∣cosθ

$$\Rightarrow 6=2\sqrt{2}\times 3\mathrm{cos}\theta$$⇒6=22​×3cosθ

$$\Rightarrow \mathrm{cos}\theta =\frac{1}{\sqrt{2}}$$⇒cosθ=2​1​

$$\Rightarrow \theta =\frac{\pi }{4}$$⇒θ=4π​

Now finding, $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\mathrm{sin}\theta =2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}=6$∣a×b∣=∣a∣∣b∣sinθ=22​×3×2​1​=6

Hence, ordered pair $(\theta ,|\vec{a}\times \vec{b}|)=(\frac{\pi }{4},6)$(θ,∣a×b∣)=(4π​,6)