Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11c

Practicing the AP Board Solutions Class 11 Maths and Chapter 11 Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11c

Introduction to 3D Geometry Exercise 11c Solutions

Introduction to 3D Geometry Class 11 Exercise 11c Solutions – Introduction to 3D Geometry 11c Exercise Solutions

I.

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:
Let D = (x, y, z) be the fourth vertex of the parallelogram.
Then the midpoint of AC = the midpoint of BD
⇒ \(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)\) = \(\left(\frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2}\right)\)
⇒ (1, 0, 2) = \(\left(\frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2}\right)\)
⇒ x = 1, y = -2 and z = 8
∴ D = (1, -2, 8)

Question 2.
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:
Given origin = Centroid of triangle PQR
⇒ (0, 0, 0) = \(\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)\)
⇒ (0, 0, 0) = \(\left(\frac{2 \mathrm{a}+4}{3}, \frac{3 \mathrm{~b}+16}{3}, \frac{2 \mathrm{c}-4}{3}\right)\)
⇒ a = -2, b = \(-\frac {16}{3}\), c = 2

Question 3.
If A and B are the points (3, 4, 5) and (-1, 3, -7), respectively, find the equation of the set of points P such that PA2 – PB2 = k2, where k is a constant.
Solution:
Let P = (x, y, z).
Then PA² + PB² = k²
⇒ (x – 3)² + (y – 4)² + (z – 5)² + (x + 1)² + (y – 3)² + (z + 7)² = k²
⇒ x² – 6x + 9 + y² – 8y + 16 + z² – 10z + 25 + x² + 2x + 1 + y² – 6y + 9 + z² + 14z + 49 = k²
⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z + 109 = k²
⇒ 2(x² + y² + z² – 2x – 7y + 2z) = k² – 109
⇒ x² + y² + z² – 2x – 7y + 2z = \(\frac{\mathrm{K}^2-109}{2}\)
Thus, the required equation is x² + y² + z² – 2x – 7y + 2z = \(\frac{\mathrm{K}^2-109}{2}\)

II.

Question 1.
Find the lengths of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0), and C(6, 0, 0).
Solution:
Let D = Midpoint of BC
= \(\left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right)\)
= (3, 2, 0)
∴ Length of Median AD = \(\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}\)
= \(\sqrt{9+4+36}\)
= √49
= 7
Let E = Midpoint of CA
= \(\left(\frac{6+0}{2}, \frac{0+0}{2}, \frac{0+6}{2}\right)\)
= (3, 0, 3)
∴ Length of Median BE = \(\sqrt{(0-3)^2+(4-0)^2+(0-3)^2}\)
= \(\sqrt{9+16+9}\)
= √34
Let F = Midpoint of AB
= \(\left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right)\)
= (0, 2, 3)
∴ Length of Median CF = \(\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}\)
= \(\sqrt{36+4+9}\)
= √49
= 7

Question 2.
A(5, 4, 6) B(1, -1, 3) and C(4, 3, 2) are three points. Find the coordinates of the point at which the bisector ∠BAC meets the side \(\overline{\mathbf{B C}}\).
Solution:
We know that the bisector of ∠BAC divides \(\overline{\mathbf{B C}}\) in the ratio AB : AC.
AB = \(\sqrt{(5-1)^2+(4+1)^2+(6-3)^2}\) = 5√2
AC = \(\sqrt{(5-4)^2+(4-3)^2+(6-2)^2}\) = 3√2
∴ AB : AC = 5 : 3
If D is the point where the bisector of ∠BAC divides \(\overline{\mathrm{BC}}\), then D divides \(\overline{\mathrm{BC}}\) in the ratio 5 : 3.
∴ D = (\(\frac{5 \times 4+3 \times 1}{5+3}\), \(\frac{5 \times 3+3 \times(-1)}{5+3}\), \(\frac{5 \times 2+3 \times 3}{5+3}\)) = \(\left(\frac{23}{8}, \frac{3}{2}, \frac{19}{8}\right)\)

Question 3.
Show that the points O(0, 0, 0), A(2, -3, 3), and B(-2, 3, -3) are collinear. Find the ratio in which each point divides the segment joining the other two points.
Solution:
OA = \(\sqrt{(0-2)^2+(0+3)^2+(0-3)^2}\)
= \(\sqrt{4+9+9}\)
= √22
AB = \(\sqrt{(2+2)^2+(-3-3)^2+(3+3)^2}\)
= \(\sqrt{16+36+36}\)
= √88
= 2√22
OB = \(\sqrt{(0+2)^2+(0-3)^2+(0+3)^2}\)
= \(\sqrt{4+9+9}\)
= √22
OA + OB = √22 + √22
= 2√22
= AB
⇒ A, O, B are collinear.
O divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 1,
A divides \(\overline{\mathrm{OB}}\) in the ratio -1 : 2,
B divides \(\overline{\mathrm{AO}}\) is in the ratio -2 : 1.