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Chapter 16: Problem 1575

For points \(\mathrm{A}(1,2,3), \mathrm{B}(5,4,1)\), the equation of plane which is perpendicular bisector of \(\underline{A B}\) is (A) \(x+2 y-7+z=0\) (B) \(2 x+y-z=7\) (C) \(x+2 y+z+7=0\) (D) \(2 x-2 y-z=7\)

Short Answer

Expert verified
The equation of the plane which is the perpendicular bisector of the line segment AB is (B) \(2x + y - z = 7\).

Step by step solution

01

Find the midpoint of the line segment AB

To find the midpoint M of the line segment AB, use the midpoint formula: \(M(x,y,z) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2}\right)\) For points A(1,2,3) and B(5,4,1): \(M(x,y,z) = \left(\frac{1 + 5}{2}, \frac{2 + 4}{2}, \frac{3 + 1}{2}\right) = (3, 3, 2)\)
02

Find the direction vector of the line segment AB

To find the direction vector of the line segment AB, subtract the coordinates of point A from point B: \(\vec{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle\) For points A(1,2,3) and B(5,4,1): \(\vec{AB} = \langle 5 - 1, 4 - 2, 1 - 3 \rangle = \langle 4, 2, -2 \rangle\) This direction vector \(\langle 4, 2, -2 \rangle\) will be normal to the plane.
03

Find the equation of the plane

Using the point-normal form of the equation of a plane: \(A(x - x_M) + B(y - y_M) + C(z - z_M) = 0\) Where (A, B, C) are the components of the normal vector, and M is the midpoint of the line segment AB. Using the normal vector \(\langle 4, 2, -2 \rangle\) and midpoint M(3, 3, 2): \(4(x - 3) + 2(y - 3) - 2(z - 2) = 0\) Expanding and simplifying, we get: \(4x + 2y - 2z = 12\) So the equation of the plane is: \(2x + y - z = 6\) Comparing this equation to the given options, we can see that it is equivalent to option (B) multiplied by a constant (-2), which means option (B) is the correct answer: Answer: (B) \(2x + y - z = 7\)

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Most popular questions from this chapter

The unit vector which is perpendicular to \((2,-4,3)\) and \((5,0,1)\), is (A) \([\\{4 / \sqrt{(585)\\}},\\{13 / \sqrt{(585)\\},\\{20 / \sqrt{(585})\\}]}\) (B) \([\\{(-4) / \sqrt{(585)\\}},\\{13 / \sqrt{(585)\\}},\\{(-20) / \sqrt{(585)\\}}]\) (C) \([\\{(-4) / \sqrt{(585)\\}},\\{(-13) / \sqrt{(585)\\},\\{20 / \sqrt{(585})\\}]}\) (D) \([\\{(-4) / \sqrt{(585)\\},\\{13 / \sqrt{(585})\\},\\{20 / \sqrt{(585})\\}]}\)

The equation of plane passing through \((1,6,-4)\) and containing \([(x-1) / 2]=[(y-2) /(-3)]=[(z-3) /(-1)]\) is (A) \(25 \mathrm{x}+14 \mathrm{y}+8 \mathrm{z}=77\) (B) \(25 \mathrm{x}+14 \mathrm{y}-8 \mathrm{y}=77\) (C) \(25 \mathrm{x}-14 \mathrm{y}-8 \mathrm{z}=77\) (D) \(25 x+14 y+8 y=-77\)

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If three vertices of rhombus are \((6,0,1)(8,-3,7)\) \((2,-5,10)\), then forth vertices \(=\) (A) \((0,-2,-4)\) (B) \((0,-2,4)\) (C) \((0,2,4)\) (D) \((0,2,-4)\)
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