** ****Distance Formula: **

**The distance between point A(x _{1}, y_{1}, z_{1}) and point B(x_{2}, y_{2}, z_{2}) is given by:**

**AB = (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√**

** **

**Q.1: Find the distance between two points whose coordinates are given below:**

**(i). (2, 8, 9) and (4, 5, 8)**

**(ii). (-3, 4, 5) and (2, 6, -1)**

**(iii). (-6, -4, 1) and (5, -2, 6)**

**(iv). (-1, 9, 8) and (6, 5, -3)**

**Sol.**

**The distance between point A (x _{1}, y_{1}, z_{1}) and point B (x_{2}, y_{2}, z_{2}) is given by:**

**AB = (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√**

**(i). The distance between point A (2, 8, 9) and point B (4, 5, 8) is:**

**Therefore, The distance between point A (2, 8, 9) and point B (4, 5, 8) is 14−−√ units**

**(ii).** **The distance between point A (****-3, 4, 5****) and point B (****2, 6, -1****) is:**

**Therefore, The distance between point A ****(****-3, 4, 5) and point B (2, 6, -1) is 65−−√ units**

** **

**(iii). The distance between point A ****(-6, -4, 1****) and point B (****5, -2, 6****) is:**

**Therefore, The distance between point A ****(****-6, -4, 1) and point B (5, -2, 6) is 56–√ units**

** **

**(iv). The distance between point A ****(-1, 9, 8****) and point B (****6, 5, -3****) is:**

**Therefore, The distance between point A ****(****-1, 9, 8) and point B (6, 5, -3) is 186−−−√ units**

** **

**Q.2: Show that the points (7, 0, -1), (-2, 3, 5) and (1, 2, 3) are collinear.**

**Sol.**

Let, the coordinates of **points A, B and C** are **(7, 0, -1), (-2, 3, 5) and (1, 2, 3)** respectively.

**Points A, B and C are collinear if they lie on the same line.**

Now,

**Therefore, AB = 314−−√ units**

Now,

**Therefore, BC = 14−−√ units**

And,

**Therefore, AC = 214−−√ units**

Since, **AC + BC = 214−−√+14−−√ = AB**

**Therefore, the points (7, 0, -1), (-2, 3, 5) and (1, 2, 3) are collinear.**

** **

**Q.3: Prove the following statements:**

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

**(ii). (4, 9, -6), (0, 7, -10) and (1, 6, -6) are the vertices of an isosceles triangle.**

**(****iii). (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram.**

** **

**Sol.**

**The distance between points A (x _{1}, y_{1}, z_{1}) and B (x_{2}, y_{2}, z_{2}) is given by:**

**AB = (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√**

** **

**(i).** Let, the **coordinates** of points **A, B and C** are **(-4, 9, 6), (0, 7, 10) and (-1, 6, 6)** respectively.

Now,

**Therefore, AB = 6 units.**

Now,

**Therefore, BC = 32–√ units.**

And,

**Therefore, AC = 32–√units.**

Now, AB^{2} = 36, BC^{2 }= 18 and AC^{2} = 18

Since, **AB ^{2} = AC^{2} + BC^{2}**. Therefore, by

**Pythagoras**

**theorem, ABC is a right angled triangle.**

**Hence, the given points (-4, 9, 6), (0, 7, 10) and (-1, 6, 6) are the vertices of a right angled triangle.**

**(ii).** Let, the **coordinates** of points **A, B and C** are **(4, 9, -6), (0, 7, -10) and (1, 6, -6) **respectively.

Now,

**Therefore, AB = 6 units.**

Now,

**Therefore, BC = 32–√ units.**

And,

**Therefore, AC = 32–√ units.**

**i.e. AC = BC ****≠**** AB**

Now, a triangle is said to be an **isosceles** triangle if **two sides** of a triangle are **equal **to each other.

Since, **AC = BC**. **Therefore triangle ABC is an isosceles triangle.**

**Hence, the given points (4, 9, -6), (0, 7, -10) and (1, 6, -6) are the vertices of an isosceles triangle.**

**(iii).** Let, the **coordinates** of **points A, B, C and D** are **(1, -2, 5), (4, -7, 8), (2, -3, 4) and (-1, 2, 1) respectively**.

Now,

**Therefore, AB = 43−−√ units.**

Now,

**Therefore, BC = 6 units.**

Now,

**Therefore, CD = 43−−√ units.**

And,

**Therefore, AD = 6 units**.

Since**, BC = AD = 6 units** and **AB = CD = 43−−√ units**

Since, it is proved that in **quadrilateral** **ABCD** **opposite** **sides** are **equal**. Therefore,** ABCD is a parallelogram.**

**Hence, the given points (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram.**

** **

**Q.4: Find the equation of the set of points P which are equidistant from point A (3, 2, 1) and point B (-1, 2, 3).**

**Sol.**

Let, **(x, y, z)** be the **coordinates** of **point P**, which is **equidistant** from **point A (3, 2, 1)** and **point B (-1, 2, 3).**

Now, according to the given condition:

**AP ^{2} = BP^{2}**

**Therefore, the required equation is: 2x – z = 0**

** **

**Q.5: The sum of distance of point P from point A (3, 0, 0) and point B (-3, 0, 0) is equal to 12 units. Find the equation of the set of points.**

**Sol.**

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

Now, according to the given condition:

**AP + BP = 12**

Now, **on squaring both the side**, we will get:

Again, **on squaring both the side**, we will get:

**Therefore, the required equation is: 3x2+4y2+4z2−108=0**

** **

**Section Formula:**

The **coordinates** of **point P**, which divides the line segment joining two points **A (a _{1}, b_{1}, c_{1})** and

**B (a**

_{2}, b_{2}, c_{2})**internally**in the ratio

**m and n**are:

The **coordinates** of **point Q**, which divides the line segment joining two points A (a_{1}, b_{1}, c_{1}) and B (a_{2}, b_{2}, c_{2}) **Externally** in the ratio **m and n** are: