NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.2


Chapter 12 Ex.12.2 Question 1

Find the distance between the following pairs of points:

(i) \(\left( {2,3,5} \right)\) and \(\left( {4,3,1} \right)\)

(ii) \(\left( { - 3,7,2} \right)\) and \(\left( {2,4, - 1} \right)\)

(iii) \(\left( { - 1,3, - 4} \right)\) and \(\left( {1, - 3,4} \right)\)

(iv) \(\left( {2, - 1,3} \right)\) and \(\left( { - 2,1,3} \right)\)

 

Solution

 

(i) \(\left( {2,3,5} \right)\) and \(\left( {4,3,1} \right)\)

Let \(P\) be \(\left( {2,3,5} \right)\) and \(Q\) be \(\left( {4,3,1} \right)\)

By using the formula,

Distance \( = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= 2,\;{y_1} = 3,\;{z_1} = 5\\{x_2} &= 4,\;{y_2} = 3,\;{z_2} = 1\end{align}\]

\[\begin{align}PQ &= \sqrt {{\left(4 - 2\right)}^2 + {{\left(3 - 3\right)}^2} + {{\left(1 - 5\right)}^2}} \\&= \sqrt {2^2 + 0^2 + {{\left(- 4 \right)}^2}} \\&= \sqrt {4 + 16} \\&= \sqrt {20} \\&= 2\sqrt 5\end{align}\]

Therefore, the required distance is \(2\sqrt 5 \) units.

(ii) \(\left( { - 3,7,2} \right)\) and \(\left( {2,4, - 1} \right)\)

Let \(P\) be \(\left( { - 3,7,2} \right)\)and \(Q\) be \(\left( {2,4, - 1} \right)\)

By using the formula,

Distance \( = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= - 3,\;{y_1} = 7,\;{z_1} = 2\\{x_2} &= 2,\;{y_2} = 4,\;{z_2} = - 1\end{align}\]

\[\begin{align}PQ &= \sqrt {{{\left( {2 - \left( { - 3} \right)} \right)}^2} + {{\left( {4 - 7} \right)}^2} + {{\left( { - 1 - 2} \right)}^2}} \\&= \sqrt {{5^2} + {{\left( { - 3} \right)}^2} + {\left( - 3 \right)}^2} \\&= \sqrt {25 + 9 + 9} \\&= \sqrt {43}\end{align}\]

Therefore, the required distance is \(\sqrt {43} \) units.

(iii) \(\left( 2,-1,3 \right)\) and \(\left( -2,1,3 \right)\)

Let \(P\) be \(\left( -1,3,-4 \right)\) and \(Q\) be \(\left( 1,-3,4 \right)\)

By using the formula,

Distance \( = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= - 1,\;{y_1} = 3,\;{z_1} = - 4\\{x_2} &= 1,\;{y_2} = - 3,\;{z_2} = 4\end{align}\]

\[\begin{align}PQ &= \sqrt {{{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left( { - 3 - 3} \right)}^2} + {{\left( {4 - \left( { - 4} \right)} \right)}^2}} \\&= \sqrt {2^2 + {{\left( - 6 \right)}^2} + {8^2}} \\&= \sqrt {4 + 36 + 64} \\&= \sqrt {104} \\&= 2\sqrt {26}\end{align}\]

Therefore, the required distance is \(2\sqrt {26} \) units.

(iv) \(\left( {2,\; - 1,\;3} \right)\;{\rm{and}}\;\left( { - 2,\;1,\;3} \right)\)

Let \(P\) be \(\left( {2,\; - 1,\;3} \right)\) and \(Q\) be \(\left( { - 2,\;1,\;3} \right)\)

By using the formula,

Distance \( = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= 2,\;{y_1} = - 1,\;{z_1} = 3\\{x_2} &= - 2,\;{y_2} = 1,\;{z_2} = 3\end{align}\]

\[\begin{align}PQ &= \sqrt {{{\left(- 2 - 2\right)}^2} + {{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left(3 - 3\right)}^2}} \\&= \sqrt {{{\left( - 4\right)}^2} + 2^2 + 0^2} \\&= \sqrt {16 + 4} \\&= \sqrt {20} \\&= 2\sqrt 5\end{align}\]

Therefore, the required distance is \(2\sqrt 5 \) units.

Chapter 12 Ex.12.2 Question 2

Show that the points \(\left( { - 2,3,5} \right),\left( {1,2,3} \right)\) and \(\left( {7,0, - 1} \right)\) are collinear.

 

Solution

 

If three points are collinear, then they lie on a line.

Firstly, let us calculate distance between the \(3\) points i.e. \(PQ,\, QR\) and \(PR.\)

Calculating PQ:

\(P = \left( { - 2,3,5} \right)\) and \(Q = \left( {1,2,3} \right)\)

By using the formula,

Distance \(PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= - 2,\;{y_1} = 3,\;{z_1} = 5\\{x_2} &= 1,\;{y_2} = 2,\;{z_2} = 3\end{align}\]

\[\begin{align}PQ &= \sqrt {\left[ {{{\left( {1 - \left( { - 2} \right)} \right)}^2} + {{\left( {2 - 3} \right)}^2} + {{\left(3 - 5\right)}^2}} \right]} \\&= \sqrt {{{\left( 3 \right)}^2} + {{\left(- 1\right)}^2} + {{\left( - 2\right)}^2}} \\&= \sqrt {9 + 1 + 4} \\&= \sqrt {14}\end{align}\]

Calculating QR:

\(Q = \left( {1,2,3} \right)\) and \(R = \left( {7,0, - 1} \right)\)

By using the formula,

Distance \(QR = \sqrt {{{\left( {x_2 - x_1} \right)}^2} + {{\left( {y_2 - y_1} \right)}^2} + {{\left( {z_2 - z_1} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= 1,\;{y_1} = 2,\;{z_1} = 3\\{x_2} &= 7,\;{y_2} = 0,\;{z_2} = - 1\end{align}\]

\[\begin{align}QR &= \sqrt {{{\left(7 - 1\right)}^2} + {{\left(0 - 2\right)}^2} + {{\left(- 1 - 3\right)}^2}} \\&= \sqrt {{6^2} + {{\left( - 2 \right)}^2} + {{\left(- 4\right)}^2}} \\&= \sqrt {36 + 4 + 16} \\&= \sqrt {56} \\&= 2\sqrt {14}\end{align}\]

Calculating PR:

\(\) and \(R = \left( {7,0, - 1} \right)\)

By using the formula,

Distance \(PR = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\[\begin{align}{x_1} &= - 2,\;{y_1} = 3,\;{z_1} = 5\\{x_2} &= 7,\;{y_2} = 0,\;{z_2} = - 1\end{align}\]

\[\begin{align}PR & = \sqrt {{{\left( {7 - \left( { - 2} \right)} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}} \\&= \sqrt {{9^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 6} \right)}^2}} \\&= \sqrt {81 + 9 + 36} \\&= \sqrt {126} \\&= 3\sqrt {14}\end{align}\]

Thus, \(PQ = \sqrt {14} \), \(QR = 2\sqrt {14} \) and \(PR = 3\sqrt {14} \)

So,

\[\begin{align}PQ + QR &= \sqrt {14} + 2\sqrt {14} \\&= 3\sqrt {14} \\&= PR\end{align}\]

Therefore, the points \(P, \,Q\) and \(R\) are collinear.

Chapter 12 Ex.12.2 Question 3

Verify the following:

(i) \(\left( 0,7, - 10 \right),\left( 1,6, - 6 \right)\) and \(\left( 4,9, - 6\right)\) are the vertices of an isosceles triangle.

(ii) \(\left( 0,7,10 \right),\left( - 1,6,6 \right)\) and \(\left( - 4,9,6 \right)\) are the vertices of a right angled triangle.

(iii) \(\left( - 1,2,1 \right),\left( 1, - 2,5 \right),\left( 4, - 7,8 \right)\) and \(\left( {2, - 3,4} \right)\) are the vertices of a parallelogram.

 

Solution

 

(i) \(\left( {0,7, - 10} \right),\left( {1,6, - 6} \right)\) and \(\left( {4,9, - 6} \right)\) are the vertices of an isosceles triangle.

Let us consider the points be \(P\left( {0,7, - 10} \right),\;Q\left( {1,6, - 6} \right)\) and \(R\left( {4,9, - 6} \right)\)

If any \(2\) sides are equal, hence it will be an isosceles triangle.

So, firstly let us calculate the length of the sides.

Calculating PQ:

\(P\left( {0,7, - 10} \right)\) and \(Q\left( {1,6, - 6} \right)\)

By using the formula,

Distance \(PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 0,\;{y_1} = 7,\;{z_1} = - 10\\{x_2} = 1,\;{y_2} = 6,\;{z_2} = - 6\end{array}\)

\[\begin{align}PQ &= \sqrt {{{\left( {1 - 0} \right)}^2} + {{\left( {6 - 7} \right)}^2} + {{\left( { - 6 - \left( { - 10} \right)} \right)}^2}} \\&= \sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {4^2}} \\&= \sqrt {1 + 1 + 16} \\&= \sqrt {18}\end{align}\]

Calculating QR:

\(Q\left( {1,6, - 6} \right)\) and \(R\left( {4,9, - 6} \right)\)

By using the formula,

Distance \(QR = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 1,\;{y_1} = 6,\;{z_1} = - 6\\{x_2} = 4,\;{y_2} = 9,\;{z_2} = - 6\end{array}\)

\[\begin{align}QR &= \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( { - 6 - \left( { - 6} \right)} \right)}^2}} \\&= \sqrt {{3^2} + {3^2} + {0^2}} \\&= \sqrt {9 + 9} \\&= \sqrt {18}\end{align}\]

Here,\(PQ = QR = \sqrt {18} \)

Since, two sides are equal, \(\Delta PQR\) is an isosceles triangle.

Thus, \(\left( {0,7, - 10} \right),\left( {1,6, - 6} \right)\) and \(\left( {4,9, - 6} \right)\) are the vertices of an isosceles triangle.

(ii) \(\left( {0,7,10} \right),\left( { - 1,6,6} \right)\) and \(\left( { - 4,9,6} \right)\) are the vertices of a right angled triangle.

Let the points be \(P\left( {0,7,10} \right),\;Q\left( { - 1,6,6} \right)\) and \(R\left( { - 4,9,6} \right)\)

Firstly, let us calculate the length of sides PQ, QR and PR.

Calculating PQ:

\(P\left( {0,7,10} \right)\) and \(Q\left( { - 1,6,6} \right)\)

By using the formula,

Distance \(PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 0,\;{y_1} = 7,\;{z_1} = 10\\{x_2} = - 1,\;{y_2} = 6,\;{z_2} = 6\end{array}\)

\[\begin{align}PQ &= \sqrt {{{\left( { - 1 - 0} \right)}^2} + {{\left( {6 - 7} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \\&= \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 4} \right)}^2}} \\&= \sqrt {1 + 1 + 16} \\&= \sqrt {18}\end{align}\]

Calculating QR:

\(Q\left( { - 1,6,6} \right)\) and \(R\left( { - 4,9,6} \right)\)

By using the formula,

Distance \(QR = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = - 1,\;{y_1} = 6,\;{z_1} = 6\\{x_2} = - 4,\;{y_2} = 9,\;{z_2} = 6\end{array}\)

\[\begin{align}QR &= \sqrt {{{\left( { - 4 - \left( { - 1} \right)} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( {6 - 6} \right)}^2}} \\&= \sqrt {{{\left( { - 3} \right)}^2} + {3^2} + {0^2}} \\&= \sqrt {9 + 9} \\&= \sqrt {18}\end{align}\]

Calculating PR:

\(P\left( {0,7,10} \right)\) and \(R\left( { - 4,9,6} \right)\)

By using the formula,

Distance \(PR = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 0,\;{y_1} = 7,\;{z_1} = 10\\{x_2} = - 4,\;{y_2} = 9,\;{z_2} = 6\end{array}\)

\[\begin{align}PR &= \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {9 - 7} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \\&= \sqrt {{{\left( { - 4} \right)}^2} + {2^2} + {{\left( { - 4} \right)}^2}} \\&= \sqrt {16 + 4 + 16} \\&= \sqrt {36}\end{align}\]

Now,

\[\begin{align}P{Q^2} + Q{R^2}& = 18 + 18\\&= 36\\&= P{R^2}\end{align}\]

By using Converse of Pythagoras theorem,

The given vertices \(P, Q\) and \(R\) are the vertices of a right-angled triangle at \(Q.\)

Thus, \(\left( {0,7,10} \right),\left( { - 1,6,6} \right)\) and \(\left( { - 4,9,6} \right)\) are the vertices of a right angled triangle.

(iii) \(\left( { - 1,2,1} \right),\left( {1, - 2,5} \right),\left( {4, - 7,8} \right)\) and \(\left( {2, - 3,4} \right)\) are the vertices of a parallelogram.

Let the points be \(A\left( { - 1,2,1} \right),\;B\left( {1, - 2,5} \right),\;C\left( {4, - 7,8} \right)\) and \(D\left( {2, - 3,4} \right)\)

if pairs of opposite sides are equal then only ABCD can be a parallelogram.

i.e., \(AB=~CD\) and \(BC=~AD\).

Firstly, let us calculate the lengths of the sides

Calculating AB:

\(A\left( { - 1,2,1} \right)\) and \(B\left( {1, - 2,5} \right)\)

By using the distance formula,

Distance \(AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = - 1,\;{y_1} = 2,\;{z_1} = 1\\{x_2} = 1,\;{y_2} = - 2,\;{z_2} = 5\end{array}\)

\[\begin{align}AB &= \sqrt {{{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left( { - 2 - 2} \right)}^2} + {{\left( {5 - 1} \right)}^2}} \\&= \sqrt {{2^2} + {{\left( { - 4} \right)}^2} + {4^2}} \\&= \sqrt {4 + 16 + 16} \\&= \sqrt {36} \\&= 6\end{align}\]

Calculating BC:

\(B\left( {1, - 2,5} \right)\) and \(C\left( {4, - 7,8} \right)\)

By using the distance formula,

Distance \(BC = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 1,\;{y_1} = - 2,\;{z_1} = 5\\{x_2} = 4,\;{y_2} = - 7,\;{z_2} = 8\end{array}\)

\[\begin{align}BC &= \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( { - 7 - \left( { - 2} \right)} \right)}^2} + {{\left( {8 - 5} \right)}^2}} \\&= \sqrt {{3^2} + {{\left( { - 5} \right)}^2} + {3^2}} \\&= \sqrt {9 + 25 + 9} \\&= \sqrt {43}\end{align}\]

Calculating CD:

\(C\left( {4, - 7,8} \right)\) and \(D\left( {2, - 3,4} \right)\)

By using the distance formula,

Distance \(CD = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 4,\;{y_1} = - 7,\;{z_1} = 8\\{x_2} = 2,\;{y_2} = - 3,\;{z_2} = 4\end{array}\)

\[\begin{align}CD &= \sqrt {\left[ {{{\left( {2 - 4} \right)}^2} + {{\left( { - 3 - \left( { - 7} \right)} \right)}^2} + {{\left( {4 - 8} \right)}^2}} \right]} \\&= \sqrt {{{\left( { - 2} \right)}^2} + {4^2} + {{\left( { - 4} \right)}^2}} \\&= \sqrt {4 + 16 + 16} \\&= \sqrt {36} \\&= 6\end{align}\]

Calculating DA:

\(D\left( {2, - 3,4} \right)\) and \(A\left( { - 1,2,1} \right)\)

By using the formula,

Distance \(DA = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = 2,\;{y_1} = - 3,\;{z_1} = 4\\{x_2} = - 1,\;{y_2} = 2,\;{z_2} = 1\end{array}\)

\[\begin{align}DA &= \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( {2 - \left( { - 3} \right)} \right)}^2} + {{\left( {1 - 4} \right)}^2}} \\&= \sqrt {{{\left( { - 3} \right)}^2} + {5^2} + {{\left( { - 3} \right)}^2}} \\&= \sqrt {9 + 25 + 9} \\&= \sqrt {43}\end{align}\]

Since, in quadrilateral \(ABCD\) both the pairs of opposite sides are equal i.e., \(AB=~CD\) and \(BC=~AD\), \(ABCD\) is a parallelogram.

Thus, \(\left( { - 1,2,1} \right),\left( {1, - 2,5} \right),\left( {4, - 7,8} \right)\) and \(\left( {2, - 3,4} \right)\) are the vertices of a parallelogram.

Chapter 12 Ex.12.2 Question 4

Find the equation of the set of points which are equidistant from the points \(\left( {1,2,3} \right)\) and \(\left( {3,2, - 1} \right)\).

 

Solution

 

Let \(A\left( {1,2,3} \right)\) and \(B\left( {3,2, - 1} \right)\)

Let point \(P\) be \(\left( {x,y,z} \right)\)

Since it is given that point \(P\left( {x,y,z} \right)\) is equidistant from the points \(A\left( {1,2,3} \right)\) and \(B\left( {3,2, - 1} \right)\)

i.e., \(PA = PB\)

Firstly, let us calculate distances \(PA\) and \(PB\)

Calculating PA:

\(P\left( {x,y,z} \right)\) and \(A\left( {1,2,3} \right)\)

By using the distance formula,

Distance \(PA = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = x,\;{y_1} = y,\;{z_1} = z\\{x_2} = 1,\;{y_2} = 2,\;{z_2} = 3\end{array}\)

\[PA = \sqrt {{{\left( {1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3 - z} \right)}^2}} \]

Calculating PB:

\(P\left( {x,y,z} \right)\) and \(B\left( {3,2, - 1} \right)\)

By using the distance formula,

Distance \(PB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{array}{l}{x_1} = x,\;{y_1} = y,\;{z_1} = z\\{x_2} = 3,\;{y_2} = 2,\;{z_2} = - 1\end{array}\)

\[PB = \sqrt {{{\left( {3 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( { - 1 - z} \right)}^2}} \]

Since, \(PA = PB\)

On squaring both the sides, we get

\[P{A^2} = P{B^2}\]

Therefore,

\[\begin{align}&{\left( {1 - x} \right)^2} + {\left( {2 - y} \right)^2} + {\left( {3 - z} \right)^2} = {\left( {3 - x} \right)^2} + {\left( {2 - y} \right)^2} + {\left( { - 1 - z} \right)^2}\\&\left( {1 + {x^2} - 2x} \right) + \left( {4 + {y^2} - 4y} \right) + \left( {9 + {z^2} - 6z} \right) = \left( {9 + {x^2} - 6x} \right) + \left( {4 + {y^2} - 4y} \right) + \left( {1 + {z^2} + 2z} \right)\\&- 2x - 4y - 6z + 14 = - 6x - 4y + 2z + 14\\&4x - 8z = 0\\&x - 2z = 0\end{align}\]

Thus, the required equation is \(x - 2z = 0\)

Chapter 12 Ex.12.2 Question 5

Find the equation of the set of points \(P,\) the sum of whose distances from \(A\left( {4,0,0} \right)\) and \(B\left( { - 4,0,0} \right)\) is equal to \(10.\)

 

Solution

 

Let \(A\left( {4,0,0} \right)\) and \(B\left( { - 4,0,0} \right)\)

Let the coordinates of point \(P\) be \(\left( {x,y,z} \right)\)

Calculating PA:

\(P\left( {x,y,z} \right)\) and \(A\left( {4,0,0} \right)\)

By using the distance formula,

Distance \(PA = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{align}{x_1} &= x,\;{y_1} = y,\;{z_1} = z\\{x_2} &= 4,\;{y_2} = 0,\;{z_2} = 0\end{align}\)

Distance \(PA = \sqrt {{{\left( {4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \)

Calculating PB:

\(P\left( {x,y,z} \right)\) and \(B\left( { - 4,0,0} \right)\)

By using the distance formula,

Distance \(PB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \)

So here,

\(\begin{align}{x_1} &= x,\;{y_1} = y,\;{z_1} = z\\{x_2} &= - 4,\;{y_2} = 0,\;{z_2} = 0\end{align}\)

Distance \(PB = \sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \)

Now it is given that

\[\begin{align}&PA + PB = 10\\&PA = 10 - PB\end{align}\]

On squaring both the sides, we get

\[\begin{align}P{A^2} &= {\left( {10 - PB} \right)^2}\\P{A^2} &= 100 + P{B^2} - 20PB\end{align}\]

Therefore,

\[\begin{align}&{\left( {4 - x} \right)^2} + {\left( {0 - y} \right)^2} + {\left( {0 - z} \right)^2} = 100 + {\left( { - 4 - x} \right)^2} + {\left( {0 - y} \right)^2} + {\left( {0 - z} \right)^2} - 20PB\\&\left( {16 + {x^2} - 8x} \right) + {y^2} + {z^2} = 100 + \left( {16 + {x^2} + 8x} \right) + {y^2} + {z^2} - 20PB\\&20PB = 16x + 100\\&5PB = \left( {4x + 25} \right)\end{align}\]

On squaring both the sides again, we get

\[\begin{align}&25P{B^2} = 16{x^2} + 200x + 625\\&25\left[ {{{\left( { - 4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \right] = 16{x^2} + 200x + 625\\&25\left[ {{x^2} + {y^2} + {z^2} + 8x + 16} \right] = 16{x^2} + 200x + 625\\&25{x^2} + 25{y^2} + 25{z^2} + 200x + 400 = 16{x^2} + 200x + 625\\&9{x^2} + 25{y^2} + 25{z^2} - 225 = 0\end{align}\]

Thus, the required equation is \(9{x^2} + 25{y^2} + 25{z^2} - 225 = 0\).

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