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Find the equation of the plane. $\newline$ Find the plane that passes through the line of intersection of the planes $\newline$ $x-z=2$ and $\newline$ $y+4z=1$ and is perpendicular to the plane $\newline$ $x+y-2z=3$ $\newline$ $x+5z=-11$

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Find the length of the transverse or conjugate axes of a hyperbola

Full solution

Q. Find the equation of the plane.

\newline

Find the plane that passes through the line of intersection of the planes

\newline

x-z=2

and

\newline

y+4z=1

and is perpendicular to the plane

\newline

x+y-2z=3

\newline

x+5z=-11

Find Direction Vector: Find the direction vector of the line of intersection of the two given planes. $\newline$ The direction vector of the line of intersection is given by the cross product of the normal vectors of the two planes. $\newline$ Normal vector of plane $1$ $x-z=2$ : $\mathbf{n}_1 = \langle 1, 0, -1 \rangle$ $\newline$ Normal vector of plane $2$ $y+4z=1$ : $\mathbf{n}_2 = \langle 0, 1, 4 \rangle$ $\newline$ Cross product: \mathbf{n}_1 \times \mathbf{n}_2 = \left| \begin{array}{ccc}\(\newlinei & j & k (\newline\)1 & 0 & -1 (\newline\)0 & 1 & 4 $\newline$ \end{array} \right|\) $\newline$ Calculating the determinant, we get: $\newline$ $i(0\cdot4 - 1\cdot(-1)) - j(1\cdot4 - 0\cdot(-1)) + k(1\cdot1 - 0\cdot0)$ $\newline$ $= i(4) - j(4) + k(1)$ $\newline$ $= \langle 4, -4, 1 \rangle$
Find Point of Intersection: Find a point of intersection of the two given planes. $\newline$ To find a point on the line of intersection, we can set one of the variables to zero and solve the system of equations formed by the two planes. $\newline$ Let's set $y = 0$ , then we have: $\newline$ $x - z = 2$ $\newline$ $0 + 4z = 1$ $\newline$ From the second equation, we get $z = \frac{1}{4}$ . Plugging this into the first equation, we get: $\newline$ $x - \frac{1}{4} = 2$ $\newline$ $x = 2 + \frac{1}{4}$ $\newline$ $x = \frac{9}{4}$ $\newline$ So, one point on the line of intersection is $(\frac{9}{4}, 0, \frac{1}{4})$ .
Use Normal Vectors: Use the direction vector of the line of intersection and the normal vector of the third plane to find the normal vector of the required plane. $\newline$ The normal vector of the third plane $x+y-2z=3$ : $n_3 = \langle 1, 1, -2 \rangle$ $\newline$ The required plane must be perpendicular to this plane, so its normal vector must be parallel to $n_3$ . $\newline$ Since the required plane also contains the line of intersection, its normal vector must be orthogonal to the direction vector found in Step $1$ . $\newline$ Therefore, the normal vector of the required plane is the cross product of the direction vector and $n_3$ . $\newline$ Direction vector: $d = \langle 4, -4, 1 \rangle$ $\newline$ Cross product: $d \times n_3 = \left| \begin{array}{ccc} i & j & k \ 4 & -4 & 1 \ 1 & 1 & -2 \end{array} \right|$ $\newline$ Calculating the determinant, we get: $\newline$ $i((-4)(-2) - 1\cdot1) - j(4(-2) - 1\cdot4) + k(4\cdot1 - (-4)\cdot1)$ $\newline$ $= i(8 - 1) - j(-8 - 4) + k(4 + 4)$ $\newline$ $= i(7) - j(-12) + k(8)$ $\newline$ $= \langle 7, 12, 8 \rangle$
Write Plane Equation: Write the equation of the plane using the normal vector and the point found in Step $2$ . $\newline$ The equation of a plane is given by: $\newline$ $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ $\newline$ where $<a, b, c>$ is the normal vector of the plane and $(x_0, y_0, z_0)$ is a point on the plane. $\newline$ Using the normal vector $<7, 12, 8>$ and the point $(\frac{9}{4}, 0, \frac{1}{4})$ , we get: $\newline$ $7(x - \frac{9}{4}) + 12(y - 0) + 8(z - \frac{1}{4}) = 0$ $\newline$ Simplifying, we get: $\newline$ $7x - \frac{63}{4} + 12y + 8z - 2 = 0$ $\newline$ Multiplying through by $4$ to clear the fractions, we get: $\newline$ $28x - 63 + 48y + 32z - 8 = 0$ $\newline$ Combining like terms, we get: $\newline$ $28x + 48y + 32z - 71 = 0$

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