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Chapter 16: Problem 1474

If \(\mathrm{A}(\mathrm{a}, 1,3), \mathrm{B}(-1, \mathrm{~b}, 2), \mathrm{C}(1,0, \mathrm{c})\) are the vertices of \(\Delta \mathrm{ABC}\) whose centroid is \((2,3,5)\), then values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are respectively (A) \(10,8,6\) (B) \(6,10,8\) (C) \(8,6,10\) (D) \(6,8,10\)

Short Answer

Expert verified
The values of a, b, and c are 6, 8, and 10. Hence, the correct choice is (D) \(6,8,10\).

Step by step solution

01

Write down the given information

: We are given: Vertex A(a, 1, 3) Vertex B(-1, b, 2) Vertex C(1, 0, c) Centroid(2, 3, 5)
02

Use the property of the centroid and set up the equations

: The property of the centroid states that the centroid coordinates are the average of the coordinates of the vertices. Mathematically, this is expressed as: Centroid G = \(\frac{A + B + C}{3}\) Applying this property to the given points: \(x_G = \frac{x_A + x_B + x_C}{3}\) \(y_G = \frac{y_A + y_B + y_C}{3}\) \(z_G = \frac{z_A + z_B + z_C}{3}\) Substitute the given coordinates: \(2 = \frac{a + (-1) + 1}{3}\) \(3 = \frac{1 + b + 0}{3}\) \(5 = \frac{3 + 2 + c}{3}\)
03

Solve the equations for a, b, and c

: Solve the equations for a, b, and c: MULTIPLY both sides by 3: \(6 = a - 1 + 1\) \(9 = 1 + b\) \(15 = 3 + 2 + c\) Subtract or add as needed: \(a = 6\) \(b = 8\) \(c = 10\)
04

Identify the correct answer

: The values of a, b, and c are 6, 8, and 10. Hence, the correct choice is: (D) \(6,8,10\)

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Most popular questions from this chapter

If \(\mathrm{G}(0)\) is centroid of \(\triangle \mathrm{ABC}\), then \(\underline{\mathrm{GA}}+\underline{\mathrm{GB}}+\underline{\mathrm{GC}}=\) (A) \(\underline{0}\) (B) 0 (C) \(\underline{x}+y+\underline{z}\) (D) \([(\underline{x}+y+\underline{z}) / 3]\)

For \(\mathrm{A}(1,5,6), \mathrm{B}(3,1,2)\) and \(\mathrm{C}(4,-1,0), \mathrm{B}\) divides \(\underline{\mathrm{AC}}\) from A in ratio (A) \(\overline{-2: 3}\) (B) \(2: 3\) (C) \(2: 1\) (D) \(-2: 1\)

The equation of plane which is passing through \((2,1,3)\) and having equal \(X\) and Y-intercept and \(Z\) -intercept 14 is (A) \(11 \mathrm{x}-11 \mathrm{y}+3 \mathrm{z}=42\) (B) \(11 \mathrm{x}+11 \mathrm{y}+3 z=42\) (C) \(11 \mathrm{x}+11 \mathrm{y}-3 \mathrm{z}=42\) (D) \(11 \mathrm{x}+11 \mathrm{y}+3 \mathrm{z}+42=0\)

If \(\mathrm{m} \angle \mathrm{B}=(\pi / 2)\) in \(\Delta \mathrm{ABC}\) and \(\mathrm{P}, \mathrm{Q}\) are points of trisection of hypotenuse \(\underline{\mathrm{A}} \mathrm{C}\), then \(\mathrm{BP}^{2}+\mathrm{BQ}^{2}=\) (A) \((5 / 9) \mathrm{AC}^{2}\) (B) \((5 / 9) \mathrm{AC}\) (C) \((25 / 81) \mathrm{AC}^{2}\) (D) \((25 / 81) \mathrm{AC}\)

The Image of point \((1,3,4)\) with respect to the plane \(2 x-y+z+3=0\) is (A) \((3,5,2)\) (B) \((-3,-5,2)\) (C) \((-3,-5,-2)\) (D) \((-3,5,2)\)
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