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Let the given planes be
\n" ); document.write( "$$P_1: 4x - y + 2z = 0$$
\n" ); document.write( "$$P_2: 4x + z = 3$$\r
\n" ); document.write( "\n" ); document.write( "We want to find the vector equation of the line of intersection of these two planes.\r
\n" ); document.write( "\n" ); document.write( "**1. Find a point on the line:**
\n" ); document.write( "Let's set $z=0$ in the equation for plane $P_2$:
\n" ); document.write( "$$4x + 0 = 3 \implies x = \frac{3}{4}$$
\n" ); document.write( "Now, substitute $x = \frac{3}{4}$ and $z=0$ into the equation for plane $P_1$:
\n" ); document.write( "$$4\left(\frac{3}{4}\right) - y + 2(0) = 0 \implies 3 - y = 0 \implies y = 3$$
\n" ); document.write( "So, the point $\left(\frac{3}{4}, 3, 0\right)$ lies on the line of intersection.\r
\n" ); document.write( "\n" ); document.write( "**2. Find the direction vector of the line:**
\n" ); document.write( "The direction vector of the line is perpendicular to the normal vectors of both planes.
\n" ); document.write( "The normal vector of $P_1$ is $\vec{n_1} = \langle 4, -1, 2 \rangle$.
\n" ); document.write( "The normal vector of $P_2$ is $\vec{n_2} = \langle 4, 0, 1 \rangle$.
\n" ); document.write( "The direction vector $\vec{d}$ is the cross product of $\vec{n_1}$ and $\vec{n_2}$:
\n" ); document.write( "$$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 2 \\ 4 & 0 & 1 \end{vmatrix}$$
\n" ); document.write( "$$\vec{d} = \mathbf{i}(-1(1) - 2(0)) - \mathbf{j}(4(1) - 2(4)) + \mathbf{k}(4(0) - (-1)(4))$$
\n" ); document.write( "$$\vec{d} = \mathbf{i}(-1) - \mathbf{j}(4 - 8) + \mathbf{k}(4)$$
\n" ); document.write( "$$\vec{d} = \langle -1, 4, 4 \rangle$$\r
\n" ); document.write( "\n" ); document.write( "**3. Write the vector equation of the line:**
\n" ); document.write( "The vector equation of a line is given by $\vec{r} = \vec{a} + t\vec{d}$, where $\vec{a}$ is a point on the line and $\vec{d}$ is the direction vector.
\n" ); document.write( "Using the point $\left(\frac{3}{4}, 3, 0\right)$ as $\vec{a}$ and $\langle -1, 4, 4 \rangle$ as $\vec{d}$, we have:
\n" ); document.write( "$$\vec{r} = \left\langle \frac{3}{4}, 3, 0 \right\rangle + t\langle -1, 4, 4 \rangle$$
\n" ); document.write( "In component form:
\n" ); document.write( "$$\vec{r} = \left\langle \frac{3}{4} - t, 3 + 4t, 4t \right\rangle$$\r
\n" ); document.write( "\n" ); document.write( "**Therefore, the vector equation of the line of intersection is:**
\n" ); document.write( "$$\vec{r} = \left\langle \frac{3}{4}, 3, 0 \right\rangle + t\langle -1, 4, 4 \rangle$$
\n" ); document.write( " \n" ); document.write( "