Question 13 of 13intermediate🔧 ApplyLong Answer4 marks

If AA and BB be the points (3,4,5)(3, 4, 5) and (−1,3,−7)(−1, 3, −7), respectively, find the equation of the set of points PP such that PA2+PB2=k2PA^2 + PB^2 = k^2, where kk is a constant.

Correct Answer

The equation of the set of points PP is:

2x2+2y2+2z2−4x−14y+4z=k2−1092x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 - 109
Exercise: Miscellaneous Exercise on Chapter 11 | Q: 4 | (Page Page 8)
For More Understanding

Explanation

This question is solved using the distance formula in 3D space. The student needs to express PA2PA^2 and PB2PB^2 in terms of coordinates of PP, then apply the given condition.

The context shows Example 6 with a similar problem where PA2+PB2=2k2PA^2 + PB^2 = 2k^2, and the same method applies here with k2k^2 instead of 2k22k^2.

Solution Steps

  1. Step 1: Let the coordinates of point PP be (x,y,z)(x, y, z).

  2. Step 2: Calculate PA2PA^2 and PB2PB^2 using the distance formula:

    PA2=(x−3)2+(y−4)2+(z−5)2PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2 PB2=(x+1)2+(y−3)2+(z+7)2PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2

  3. Step 3: By the given condition PA2+PB2=k2PA^2 + PB^2 = k^2, we have:

    (x−3)2+(y−4)2+(z−5)2+(x+1)2+(y−3)2+(z+7)2=k2(x - 3)^2 + (y - 4)^2 + (z - 5)^2 + (x + 1)^2 + (y - 3)^2 + (z + 7)^2 = k^2

  4. Step 4: Expanding all terms:

    (x2−6x+9)+(y2−8y+16)+(z2−10z+25)+(x2+2x+1)+(y2−6y+9)+(z2+14z+49)=k2(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25) + (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49) = k^2

  5. Step 5: Simplifying and collecting like terms:

    2x2+2y2+2z2−4x−14y+4z+109=k22x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2

    Therefore,

    2x2+2y2+2z2−4x−14y+4z=k2−1092x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 - 109
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