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Step-by-Step Solution
Step 1: Identify the Normal Vectors of the Two Planes
Each plane is defined by two direction vectors lying in that plane:
• Plane 1: direction vectors $ \hat{i} + \hat{j} $ and $ \hat{i} + \hat{k} $.
• Plane 2: direction vectors $ \hat{i} - \hat{j} $ and $ \hat{j} - \hat{k} $.
The normal vector to a plane can be found by taking the cross product of two non-parallel vectors that lie in the plane. Let
$ \mathbf{n}_1 $ be the normal to the first plane and $ \mathbf{n}_2 $ be the normal to the second plane.
Step 2: Compute the Normal Vectors
For Plane 1:
$$
\mathbf{n}_1
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
1 & 0 & 1
\end{vmatrix}
\,.
$$
Expanding,
$$
\mathbf{n}_1
= \hat{i}\bigl(1 \cdot 1 - 0 \cdot 0\bigr)
- \hat{j}\bigl(1 \cdot 1 - 0 \cdot 1\bigr)
+ \hat{k}\bigl(1 \cdot 0 - 1 \cdot 1\bigr)
= \hat{i} - \hat{j} - \hat{k}.
$$
For Plane 2:
$$
\mathbf{n}_2
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 0 \\
1 & 0 & -1
\end{vmatrix}
\,.
$$
Expanding,
$$
\mathbf{n}_2
= \hat{i}\bigl((-1)(-1) - 0 \cdot 0\bigr)
- \hat{j}\bigl(1 \cdot (-1) - 0 \cdot 1\bigr)
+ \hat{k}\bigl(1 \cdot 0 - (-1)\cdot 1\bigr)
= \hat{i} + \hat{j} + \hat{k}.
$$
Step 3: Find a Vector Parallel to the Line of Intersection
A vector parallel to the line of intersection of the two planes is perpendicular to both $ \mathbf{n}_1 $ and $ \mathbf{n}_2 $. Such a vector can be taken as
$$
\vec{a} \;=\; \lambda \,\bigl(\mathbf{n}_1 \times \mathbf{n}_2\bigr).
$$
Any non-zero scalar multiple of this cross product will be a valid direction vector along the line of intersection.
Step 4: Calculate $ \mathbf{n}_1 \times \mathbf{n}_2 $
Let
$ \mathbf{n}_1 = (1, -1, -1) $
and
$ \mathbf{n}_2 = (1, 1, 1). $
Then:
$$
\mathbf{n}_1 \times \mathbf{n}_2
= \bigl(1, -1, -1\bigr)\,\times\,\bigl(1, 1, 1\bigr).
$$
Using the component formula for cross product:
$$
\mathbf{n}_1 \times \mathbf{n}_2
= \bigl(\,(-1)\cdot1 - (-1)\cdot1,\;\,(-1)\cdot1 - 1\cdot1,\;\,1\cdot1 - (-1)\cdot1\bigr)
= \bigl(0,\;-2,\;2\bigr).
$$
So
$$
\vec{a}
= \lambda \,\bigl(0\,\hat{i} \;-\;2\,\hat{j} \;+\;2\,\hat{k}\bigr).
$$
Step 5: Use the Condition $ \vec{a} \cdot \vec{b} = 6 $
We are given $ \vec{b} = 2\,\hat{i} - 2\,\hat{j} + \hat{k} $ and
$ \vec{a} \cdot \vec{b} = 6 $. Substituting
$ \vec{a} = \lambda\,(0\,\hat{i} - 2\,\hat{j} + 2\,\hat{k}) $:
$$
\vec{a} \cdot \vec{b}
= \lambda\,\bigl(0\,\hat{i} - 2\,\hat{j} + 2\,\hat{k}\bigr)
\cdot\bigl(2\,\hat{i} - 2\,\hat{j} + \hat{k}\bigr).
$$
Notice
$ \hat{i}\cdot\hat{j} = 0$,
$ \hat{j}\cdot\hat{k} = 0$, etc., so the dot product simplifies to:
$$
\vec{a} \cdot \vec{b}
= \lambda\,\bigl[(-2)\times(-2) + (2)\times(1)\bigr]
= \lambda\,(4 + 2)
= 6\lambda.
$$
Since this must equal 6,
$$
6\lambda = 6
\quad\Longrightarrow\quad
\lambda = 1.
$$
Thus
$$
\vec{a} = (0)\,\hat{i} - 2\,\hat{j} + 2\,\hat{k}.
$$
Step 6: Find the Angle $ \theta $ between $ \vec{a} $ and $ \vec{b} $
The angle $ \theta $ between two vectors $ \vec{u} $ and $ \vec{v} $ is given by
$$
\cos\theta
= \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\;|\vec{v}|}.
$$
We already know $ \vec{a} \cdot \vec{b} = 6 $. Next, compute the magnitudes:
$$
|\vec{a}|
= \sqrt{(0)^2 + (-2)^2 + (2)^2}
= \sqrt{4 + 4}
= \sqrt{8}
= 2\sqrt{2},
$$
$$
|\vec{b}|
= \sqrt{(2)^2 + (-2)^2 + (1)^2}
= \sqrt{4 + 4 + 1}
= \sqrt{9}
= 3.
$$
Therefore,
$$
\cos\theta
= \frac{6}{(2\sqrt{2}) \cdot 3}
= \frac{6}{6\sqrt{2}}
= \frac{1}{\sqrt{2}}
\;\;\Longrightarrow\;\;
\theta = \frac{\pi}{4}.
$$
Step 7: Compute $ |\vec{a} \times \vec{b}| $
The cross product $ \vec{a} \times \vec{b} $ can be computed via the determinant:
$$
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & -2 & 2 \\
2 & -2 & 1
\end{vmatrix}.
$$
Expanding component-wise:
• $ \hat{i} $ component: $(-2)\times1 - 2\times(-2) = -2 + 4 = 2,$
• $ \hat{j} $ component: $0\times1 - 2\times2 = -4,$ but it appears with a minus sign in the determinant expansion, resulting in $+4\,\hat{j},$
• $ \hat{k} $ component: $0\times(-2) - (-2)\times2 = 4.$
Hence,
$$
\vec{a} \times \vec{b} = 2\,\hat{i} + 4\,\hat{j} + 4\,\hat{k},
$$
and its magnitude is
$$
|\vec{a} \times \vec{b}|
= \sqrt{2^2 + 4^2 + 4^2}
= \sqrt{4 + 16 + 16}
= \sqrt{36}
= 6.
$$
Step 8: Final Answer
The ordered pair $ (\theta, \,|\vec{a} \times \vec{b}|) $ is therefore
$$
\left(\frac{\pi}{4}, \,6\right).
$$
That confirms the correct choice is
$ \bigl(\tfrac{\pi}{4}, 6\bigr).
$
