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

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 12 Introduction to three Dimensional Geometry Miscellaneous Exercise

Question 1.

Three vertices of a parallelogram ABCD are A(3, – 1, 2), B(1, 2, – 4) and C(- 1, 1, 2). Find the coordinates of the fourth vertex.

Solution:

The vertices A and C are (3, – 1, 2) and (- 1, 1, 2) respectively

âˆ´ Mid-point of AC is (\(\frac{3-1}{2}\), \(\frac{- 1+1}{2}\), \(\frac{2+2}{2}\)), i.e., (1, 0, 2).

Let the point D be (x_{1}, y_{1}, z_{1}).

Now, (1, 0, 2) is the mid-point of diagonal BD also.

âˆ´ The coordinates of point D are (1, – 2, 8).

Question 2.

Find the lengths of medians of the triangle whose vertices are A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0).

Solution:

(i) The points B and C are (0, 4, 0) and (6, 0, 0).

âˆ´ Middle point D of it is (\(\frac{0+6}{2}\), \(\frac{4+0}{2}\), \(\frac{0+0}{2}\)) or (3, 2, 0)

The point A is (0, 0, 6) and the point D is (3, 2, 0).

(ii) The mid-point E of AC, where A is (0, 0, 6) and C is (6, 0, 0) is (\(\frac{0+6}{2}\), \(\frac{0+0}{2}\), \(\frac{6+0}{2}\)), i.e; (3, 0, 3).

Now, the point B is (0, 4, 0) and E is (3, 0, 3).

âˆ´ Length of median BE = \(\sqrt{(3-0)^{2}+(0-4)^{2}+(3-0)^{2}}\)

= \(\sqrt{9+16+9}\) = \(\sqrt{34}\).

(iii) The mid-point F of the line segment joining the points A(0, 0, 6) and B(0, 4, 0) is

(\(\frac{0+0}{2}\), \(\frac{0+4}{2}\), \(\frac{6+0}{2}\)) i.e; (0, 2, 3)

Now C is (6, 0, 0) and mid-point F of AB is (0, 2, 3).

âˆ´ Length of median CF = \(\sqrt{(0-6)^{2}+(2-0)^{2}+(3-0)^{2}}\)

= \(\sqrt{36+4+9}\) = 7.

Question 3.

If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(- 4, 3b, – 10) and R(8, 14, 2c), then find the values of a, b and c.

Solution:

Vertices of âˆ†PQR are P(2a, 2, 6), Q(- 4, 3b, – 10) and R(8, 14, 2c).

âˆ´ Centroid of the âˆ† PQR is

We are given that centroid is origin i.e., (0, 0, 0).

Question 4.

Find the coordinates of a point on y-axis which is at a distance of 5\(\sqrt{2}\) from the point P(3, – 2, 5).

Solution:

A point lying on y-axis is A(0, y, 0).

The coordinates of point P are (3, – 2, 5).

But we have:

AP = 5\(\sqrt{2}\)

âˆ´ \(\sqrt{\left(y_{1}+2\right)^{2}+34}\) = 5\(\sqrt{2}\)

or (y_{1} + 2)^{2} = 50 – 34 = 16.

âˆ´ y_{1} + 2 = Â± 4.

âˆ´ y_{1} = 2, – 6

Hence, there are two points (0, 2, 0) and (0, – 6, 0), which are at a distance of 5\(\sqrt{2}\) from P(3, – 2, 5).

Question 5.

A point R with x-coordinate 4 lies on the line segment joining the points P(2, – 3, 4) and Q(8, 0, 10). Find the coordinates of the point R.

Solution:

Let R divides PQ in the ratio k : 1, where P is (2, – 3, 4) and Q is (8, 0, 10).

âˆ´ The coordinates of point R are (\(\frac{8k+2}{k+1}\), \(\frac{0-3}{k+1}\), \(\frac{10k+4}{k+1}\))

But x-coordinate of R is 4.

âˆ´ The point R is (4, – 2, 6).

Question 6.

If A and B are the points (3, 4, 5) and (- 1, 3, – 7) respectively, find the equation of set of points P such that PA^{2} + PB^{2} = k^{2}, where k is a constant.

Solution:

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

The given points are A(3, 4, 5) and B(- 1, 3, – 7).

So, PA^{2} + PB^{2} = [(x – 3)^{2} + (y – 4)^{2} + (z – 5)^{2}] + [(x + 1)^{2} + (y – 3)^{2} + (z + 7)^{2}]

= [x^{2} + y^{2} + z^{2} – 6x – 8y – 10z + 9 + 16 + 25] + [x^{2} + y^{2} + z^{2} + 2x – 6y + 14z + 1 + 9 + 49]

or 2(x^{2} + y^{2} + z^{2}) – 4x – 14y + 4z + 50 + 59 = k^{2}.

or 2(x^{2} + y^{2} + z^{2}) – 4x – 14y + 4z + 109 – k^{2} = 0.