91Ó°ÊÓ

The direction vector parallel to the line of intersection of two planes is perpendicular to both normal vectors of these planes. To find the normal vectors, we take the cross product of the vectors parallel to each plane. For plane \(P_1\), we have vectors \(\mathbf{v}_1 = 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}\) and \(\mathbf{v}_2 = 4 \hat{\mathbf{j}} - 3 \hat{\mathbf{k}}\). For plane \(P_2\), we have \(\mathbf{v}_3 = \hat{\mathbf{j}} - \hat{\mathbf{k}}\) and \(\mathbf{v}_4 = 3 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}\).

Compute \(\mathbf{n}_1 = \mathbf{v}_1 \times \mathbf{v}_2\) for plane \(P_1\): \[ \mathbf{n}_1 = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \0 & 2 & 3 \0 & 4 & -3 \end{vmatrix} = -12\hat{\mathbf{i}} + 0\hat{\mathbf{j}} + 8\hat{\mathbf{k}} = -12\hat{\mathbf{i}} + 8\hat{\mathbf{k}} \]Compute \(\mathbf{n}_2 = \mathbf{v}_3 \times \mathbf{v}_4\) for plane \(P_2\): \[ \mathbf{n}_2 = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \0 & 1 & -1 \3 & 3 & 0 \end{vmatrix} = 3\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - 3\hat{\mathbf{k}} \]

The direction vector \(\mathbf{A}\) is perpendicular to both \(\mathbf{n}_1\) and \(\mathbf{n}_2\), hence \(\mathbf{A} = \mathbf{n}_1 \times \mathbf{n}_2\):\[ \mathbf{A} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \-12 & 0 & 8 \3 & -3 & -3 \end{vmatrix} = 24\hat{\mathbf{i}} + 60\hat{\mathbf{j}} + 36\hat{\mathbf{k}} \]

The direction vector can be simplified by dividing by the greatest common factor. Simplifying \(24\hat{\mathbf{i}} + 60\hat{\mathbf{j}} + 36\hat{\mathbf{k}}\), we divide by 12:\[ \mathbf{A} = 2\hat{\mathbf{i}} + 5\hat{\mathbf{j}} + 3\hat{\mathbf{k}} \]

The given vector is \(2\hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}}\). To find the angle \(\theta\) between \(\mathbf{A} = 2\hat{\mathbf{i}} + 5\hat{\mathbf{j}} + 3\hat{\mathbf{k}}\) and this vector, use the dot product formula:\[ \cos \theta = \frac{\mathbf{A} \cdot \text{Given Vector}}{\|\mathbf{A}\| \|\text{Given Vector}\|} \]Compute the dot product:\[ \mathbf{A} \cdot (2\hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}}) = 2(2) + 5(1) + 3(-2) = 4 + 5 - 6 = 3 \]Compute magnitudes:\[ \|\mathbf{A}\| = \sqrt{2^2 + 5^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}, \|\text{Given Vector}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3 \]Compute \(\cos \theta\):\[ \cos \theta = \frac{3}{\sqrt{38} \cdot 3} = \frac{1}{\sqrt{38}} \]Since \(\cos(\theta) = \frac{1}{\sqrt{38}}\), the angle \(\theta\) is closest to \(\frac{\pi}{4}\) for practical angle options.