If (2, 3, 5) is one end of a diameter of the sphere x²+ y²+ z² − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are

(4, 9, –3)

(4, –3, 3)

(4, 3, 5)

(4, 3, –3)

Solution

(4, 9, –3)

Coordinates of centre (3, 6, 1)
Let the coordinates of the other end of diameter are (α, β, γ)
$then space fraction numerator straight alpha space plus 2 over denominator 2 end fraction space equals space 3 comma space fraction numerator straight beta space plus space 3 over denominator 2 end fraction space equals 6 comma space space fraction numerator straight gamma space plus 5 over denominator 2 end fraction space equals 1$
Hence α = 4, β = 9 and γ = −3.

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Three Dimensional Geometry

If (2, 3, 5) is one end of a diameter of the sphere x²+ y²+ z² − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are

(4, 9, –3)

(4, –3, 3)

(4, 3, 5)

(4, 3, –3)

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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Email Address

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For Daily Free Study Material Join wiredfaculty

Three Dimensional Geometry

If (2, 3, 5) is one end of a diameter of the sphere x2+ y2+ z2 − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are (4, 9, –3) (4, –3, 3) (4, 3, 5) (4, 3, –3)

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

If (2, 3, 5) is one end of a diameter of the sphere x²+ y²+ z² − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are

(4, 9, –3)

(4, –3, 3)

(4, 3, 5)

(4, 3, –3)