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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

\([(x-1) / 3]=[(y+1) / 2]=[(z-1) / 5]\) and \([(\mathrm{x}-2) / 4]=[(\mathrm{y}-1) / 3]=[(\mathrm{z}+1) /(-2)]\) lines are (A) parallel (B) coincident (C) Intersecting (D) skew

The co-ordinates of the points of trisection of \(\underline{A B}\) is where \(\mathrm{A}(-5,7,2), \mathrm{B}(1,3,7)\) (A) \([-1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (B) \([1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (C) \([-1,4,(16 / 3)][-3,\\{(-11) / 2\\},\\{(-11) / 3\\}]\) (D) None of these

The locus of point of the plane passing through \((\alpha, \beta, \gamma)\) and intersect the axis in \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and the plane which is parallel to such plane is (A) \((\mathrm{x} / \alpha)+(\mathrm{y} / \beta)+(\mathrm{z} / \gamma)=1\) (B) \((\alpha / x)+(\beta / y)+(\gamma / z)=1\) (C) \(x+y+z=1\) (D) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\alpha \beta \gamma\)

If angle between two unit vectors \(\underline{a} \& \underline{b}\) is \(\alpha\), then \(|\underline{\mathrm{a}}-\underline{\mathrm{b}} \cos \alpha|=0<\alpha<(\pi / 2)\) (A) \(\sin \alpha\) (B) \(\sin (\alpha / 2)\) (C) \(\sin 2 \alpha\) (D) \(\sin ^{2}(\alpha / 2)\)

\(\underline{\mathrm{x}}=(2,-6,3), \mathrm{y}=(1,2,-2)\) and \(\underline{\mathrm{x}}^{\wedge} \mathrm{y}=\theta\), then \(\sin \theta=\) (A) \([21 / \sqrt{(185)}]\) (B) \(-[\sqrt{(185) / 21]}\) (C) \(-[21 / \sqrt{(185)}]\) (D) \([\sqrt{(185) / 21]}\)
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