Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i}+\hat{j}, \hat{i}+\hat{k}$ and $\hat{i}-\hat{j}, \hat{j}-\hat{k}$. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$, then the ordered pair $(\theta,|\vec{a} \times \vec{b}|)$ is equal to :

We have, $\vec{a}$ is non-zero vector parallel to the line of intersection of the two planes described by $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$. <br/><br/>Let $\mathbf{n}_1$ and $\mathbf{n}_2$ are the normal vector to the plane $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$, respectively. <br/><br/>$$ \begin{aligned} & n_1=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & n_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \vec{a}=\lambda\left|n_1 \times n_2\right| \end{aligned} $$ <br/><br/>[ $\because \mathbf{a}$ is parallel to line of intersection of both planes] <br/><br/>$$ \begin{aligned} & =\lambda\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{array}\right| \\\\ & =\lambda(-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $$ <br/><br/>$\vec{a} \cdot \vec{b} =6 [Given]$ <br/><br/>$\lambda(0+4+2) =6$ <br/><br/>$\lambda =1$ <br/><br/>$$ \begin{aligned} \therefore \vec{a} & =-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ \cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{\sqrt{4+4} \sqrt{4+4+1}} \\\\ & =\frac{1}{\sqrt{2}} \end{aligned} $$ <br/><br/>$\therefore \theta=\frac{\pi}{4}$ <br/><br/>$$ \begin{aligned} \vec{a} \times \vec{b} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 2 \\ 2 & -2 & 1 \end{array}\right| \\\\ & =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\ |\vec{a} \times \vec{b}| & =\sqrt{4+16+16}=6 \\\\ \text { Hence, }(\theta,|\vec{a} \times \vec{b}|) & =\left(\frac{\pi}{4}, 6\right) \end{aligned} $$