Let A (4, 0, 0) & B (– 4, 0, 0)

Let the coordinates of point P be (x, y, z)

Calculating PA

P ≡ (x, y, z) and A ≡ (4, 0, 0)

By using the formula,

Distance PA =

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 4, y_{2} = 0, z_{2} = 0

Distance PA =

\( \sqrt{(4– x)^2 + (0 – y)^2 + (0 – z)^2}\)

Calculating PB

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

By using the formula,

Distance PB

= \( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = – 4, y_{2} = 0, z_{2} = 0

Distance PB

= \( \sqrt{(4– x)^2 + (0 – y)^2 + (0 – z)^2}\)

Now it is given that:

PA + PB = 10

PA = 10 – PB

Square on both the sides, we get

PA^{2} = (10 – PB)^{2}

PA^{2} = 100 + PB^{2} – 20 PB

(4 – x)^{2} + (0 – y)^{2} + (0 – z)^{2}

100 + (– 4 – x)^{2} + (0 – y)^{2} + (0 – z)^{2} – 20 PB

(16 + x^{2} – 8x) + (y^{2}) + (z^{2})

100 + (16 + x^{2} + 8x) + (y^{2}) + (z^{2}) – 20 PB

20 PB = 16x + 100

5 PB = (4x + 25)

Square on both the sides again, we get

25 PB^{2} = 16x^{2} + 200x + 625

25 [(– 4 – x)^{2} + (0 – y)^{2} + (0 – z)^{2}] = 16x^{2} + 200x + 625

25 [x^{2} + y^{2} + z^{2} + 8x + 16] = 16x^{2} + 200x + 625

25x^{2} + 25y^{2} + 25z^{2} + 200x + 400 = 16x^{2} + 200x + 625

9x^{2} + 25y^{2} + 25z^{2} – 225 = 0

The required equation is 9x^{2} + 25y^{2} + 25z^{2} – 225 = 0

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.