Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 12 Introduction to three Dimensional Geometry Ex 12.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 12 Introduction to three Dimensional Geometry Ex 12.2

Question 1.

Find the distance between the following pairs of points:

(i) (2, 3, 5), (4, 3, 1)

(ii) (- 3, 7, 2), (2, 4, – 1)

(iii) (-1, 3, – 4), (1, – 3, 4)

(iv) (2, – 1, 3), (- 2, 1, 3).

Solution:

(i) The given points are (2, 3, 5) and (4, 3, 1).

Question 2.

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

Solution:

Let the given points be A(- 2, 3, 5), B(0, 2, 3) and C(7, 0, – 1).

Now, AB + BC = \(\sqrt{14}\) + 2\(\sqrt{14}\) = 3\(\sqrt{14}\).

Thus, AB + BC = AC.

Hence, the points A, B and C are collinear.

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:

(i) Let A(0, 7, – 10), B(1, 6, – 6) and C(4, 9, – 6) donate the given points. Then,

∴ AB = BC

Thus, the ∆ABC is isosceles.

(ii) Let A(0, 7, 10), B(- 1, 6, 6) and C(- 4, 9, 6) denote the given points. Then,

Now, AB^{2} + BC^{2} = (3\(\sqrt{2}\))^{2} + (3\(\sqrt{2}\))^{2}

= 18 + 18 = 36.

Also, AC^{2} = 36.

∴ AB^{2} + BC^{2} = AC^{2}

⇒ ABC is right ∠d triangle with ∠B = 90°.

(iii) The given points are A(- 1, 2, 1), B(1, – 2, 5), C(4, – 7, 8) and D(2, – 3, 4). Then, by distance formula,

Now, AB = CD and BC = DA, i.e., opposite sides are equal and AC ≠ BD, i.e., the diagonals are not equal.

So, ABCD is a parallelogram.

Question 4.

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

Solution:

Let A(1, 2, 3) and B(3, 2, – 1) be the given points and the coordinates of points P be (x, y, z).

We are given AP = BP

or AP^{2} = BP^{2}

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

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

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

or – 2x – 4y – 6z = – 6x – 4y + 2z

or 4x – 8z = 0

or x – 2z = 0.

Question 5.

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

Solution:

The given points are A(4, 0, 0) and B(- 4, 0, 0).

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

We have:

PA + PB = 10.

Squaring both sides, we get

(x – 4)^{2} + y^{2} + z^{2} = 100 + (x + 4)^{2} + y^{2} + z^{2} – 20 \(\sqrt{(x+4)^{2}+y^{2}+z^{2}}\)

Squaring again, we get

16x^{2} + 200x + 625 = 25(x^{2} + y^{2} + z^{2} + 8x + 16)

or 9x^{2} + 25y^{2} + 25z^{2}

= 625 – 400

= 225

i.e., 9x^{2} + 25y^{2} + 25z^{2} = 225.