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If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.

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Question

If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.

Sum
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Solution

Given: The vertices of triangle PQR are P(2a, 2, 6), Q(−4, 3b, –10), R(8, 14, 2c).

∴ Centroid of ∆PQR `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3 - z_1 + z_2 +z_3)/3)`

or `((2a - 4 + 8)/3, (2 + 3b + 14)/3, (6 - 10 + 2c)/3)`

or `((2a + 4)/3, (3b + 16)/3, (2c - 4)/3)`

Since, the centroid is the origin (0, 0, 0), then

∴ `(2a + 4)/3 = 0,` or a = −2

`(3b + 16)/3 = 0, b = - 16/3`

`(2c - 4)/3 = 0, c = 2`

Hence, the values ​​of a, b and c are −2, `-16/3` and 2 respectively.

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Chapter 11: Introduction to Three Dimensional Geometry - Miscellaneous Exercise [Page 215]

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NCERT Mathematics [English] Class 11
Chapter 11 Introduction to Three Dimensional Geometry
Miscellaneous Exercise | Q 3. | Page 215

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