**NCERT Solutions for Class 11 Maths Chapter 11 ****Introduction to Three Dimensional Geometry**** Ex 11.2**

**Ex 11.2 Class 11 Maths Question 1.**

**Find the distance between the following pairs of points:****(i)** (2, 3, 5) and (4, 3, 1)**(ii)** (-3, 7, 2) and (2, 4, -1)**(iii)** (-1, 3, -4) and (1, -3, 4)**(iv)** (2, -1, 3) and (-2, 1, 3)

**Solution:**

### Ex 11.2 Class 11 Maths Question 2:

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

Hence,** **the points P(-2, 3, 5), Q(1, 2, 3) and R(7, 0, -1) are collinear.

**Ex 11.2 Class 11 Maths Question 3:**

**Verify the following:**

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

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

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

**Solution:**

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.

Therefore, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

**Ex 11.2 Class 11 Maths Question 4:**

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).

**Solution:**

Let P(x, y, z) be the point that is equidistant from the points A(1, 2, 3) and B(3, 2, -1).

Therefore, PA = PB

Or, PA^{2} = PB^{2}

(x – 1)^{2} + (y – 2)^{2} + (z – 3)^{2} = (x – 3)^{2} + (y – 2)^{2} + (z + 1)^{2}

x^{2} – 2x + 1 + y^{2} – 4y + 4 + z^{2} – 6z + 9 = x^{2} – 6x + 9 + y^{2} – 4y + 4 + z^{2} + 2z + 1

-2x – 4y – 6z + 14 = -6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

Thus, the required equation is x – 2z = 0.

**Ex 11.2 Class 11 Maths Question 5:**

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

**Solution:**

**Related Links:**

**NCERT Solutions for Maths Class 9**