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Question
A tetrahedron is a three dimensional figure bounded by four non-coplanar triangular planes. So, a tetrahedron has four non-coplanar points as its vertices.
Let a tetrahedron has four points A, B, C, D whose co-ordinates are (xi,yi,zi), i=1, 2, 3, 4 respectively in a rectangular three dimensional space. The co-ordinates of its centroid are given by ∑4i=1xi4,∑4i=1yi4,∑4i=1zi4. The circumcenter of the tetrahedron is the center of a sphere passing through its vertices. So this point is equidistant from each of vertices of tetrahedron.
Let tetrahedron has three of its vertices represented by the points A (6, - 5, -1), B(- 4, 1, 3) & C(2, -4, 18) & its centroid lies at the point (1, -2, 5).
On the basis of above information answer the following questions.The co-ordinate of centre of the sphere circumscribe the tetrahedron is
Let a tetrahedron has four points A, B, C, D whose co-ordinates are (xi,yi,zi), i=1, 2, 3, 4 respectively in a rectangular three dimensional space. The co-ordinates of its centroid are given by ∑4i=1xi4,∑4i=1yi4,∑4i=1zi4. The circumcenter of the tetrahedron is the center of a sphere passing through its vertices. So this point is equidistant from each of vertices of tetrahedron.
Let tetrahedron has three of its vertices represented by the points A (6, - 5, -1), B(- 4, 1, 3) & C(2, -4, 18) & its centroid lies at the point (1, -2, 5).
On the basis of above information answer the following questions.The co-ordinate of centre of the sphere circumscribe the tetrahedron is
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Solution
The correct option is C (87,−457,8)
Let P(a, b, c) be the centre of the sphere circumscribing the tetrahedron. Let r be the radius of the sphere.
∴ r2=(a−0)2+(b−0)2+(c−0)2=a2+b2+c2 (i)
(PA)2=(a−6)2+(b+5)2+(c+1)2=a2+b2+c2−12a+10b+2c+62 (ii)
(PB)2=(a+4)2+(b−1)2+(c−3)2=a2+b2+c2+8a−2b−6c+26 (iii)
(PC)2=(a−2)2+(b+4)2+(c−18)2=a2+b2+c2−4a+8b−36c+344 (iv)
From (i) & (iv) → a−2b+9c=86
From (i) & (iii) → 4a−b−3c=−13
From (i) & (ii) → 6a−5b−c=31
∴ a∣∣
∣∣−2986−1−3−13−5−131∣∣
∣∣=−b∣∣
∣∣19864−3−136−131∣∣
∣∣=c∣∣
∣∣1−2864−1−36−5−1∣∣
∣∣
=a−128=−b−720=c−896=1−112
⇒ a=128112, b=−720112, c=896112
⇒ (a,b,c)=(87,−457,8)
Let P(a, b, c) be the centre of the sphere circumscribing the tetrahedron. Let r be the radius of the sphere.
∴ r2=(a−0)2+(b−0)2+(c−0)2=a2+b2+c2 (i)
(PA)2=(a−6)2+(b+5)2+(c+1)2=a2+b2+c2−12a+10b+2c+62 (ii)
(PB)2=(a+4)2+(b−1)2+(c−3)2=a2+b2+c2+8a−2b−6c+26 (iii)
(PC)2=(a−2)2+(b+4)2+(c−18)2=a2+b2+c2−4a+8b−36c+344 (iv)
From (i) & (iv) → a−2b+9c=86
From (i) & (iii) → 4a−b−3c=−13
From (i) & (ii) → 6a−5b−c=31
∴ a∣∣ ∣∣−2986−1−3−13−5−131∣∣ ∣∣=−b∣∣ ∣∣19864−3−136−131∣∣ ∣∣=c∣∣ ∣∣1−2864−1−36−5−1∣∣ ∣∣
=a−128=−b−720=c−896=1−112
⇒ a=128112, b=−720112, c=896112
⇒ (a,b,c)=(87,−457,8)
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