Referring to the AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry Exercise 11b Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11b
I.
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:
The distance between P(x1, y1, z1) & Q(x2, y2, z2) is given by
PQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)
i) Distance between P(2, 3, 5) & Q(4, 3, 1)
= \(\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}=\sqrt{(2)^2+(0)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}\) = 2√5
ii) Distance between P(-3, 7, 2) & Q(2, 4, -1)
= \(\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}=\sqrt{(5)^2+(-3)^2+(-3)^2}=\sqrt{25+9+9}\)
iii) Distance between P(-1, 3, -4) & Q(1, -3, 4)
= \(\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}=\sqrt{(2)^2+(-6)^2+(8)^2}=\sqrt{4+36+64}=\sqrt{104}\)
iv) Distance between P(2, -1, 3) & Q(-2, 1, 3)
= \(\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2}=\sqrt{(-4)^2+(2)^2+(0)^2}=\sqrt{16+4}=\sqrt{20}\) = 2√5
Question 2.
Find the distance of P(3, -2, 4) from the origin.
Solution:
Given point is P(3, -2, 4) = (x1; y1, z1), Coordinates of origin ‘O’ = (0, 0, 0).
Distance OP = \(\sqrt{x_1^2+y_1^2+z_1^2}=\sqrt{9+4+16}=\sqrt{29}\)
Question 3.
Find ‘x’ if the distance between (5, -1, 7) and (x, 5, 1) is ‘9’ units.
Solution:
Given points are A(5, -1, 7), B(x, 5, 1)
⇒ Given that AB = 9
Squaring on both sides AB2 = 81
⇒ (x – 5)2 + (5 + 1)2 + (1 – y)2 = 81
⇒ (x – 5)2 + 36 + 36 = 81
⇒ (x – 5)2 = 81 – 72 = 9
⇒ x – 5 = ± 3
⇒ x – 5 = 3; x – 5 = -3
⇒ x = 8 (or) 2
II.
Question 1.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let the points P(-2, 3, 5), Q(1, 2, 3) & R(7, 0, -1) are collinear if they lie on a same line.
PQ = \(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}=\sqrt{(3)^2+(-1)^2+(-2)^2}=\sqrt{9+1+4}=\sqrt{14}\)
QR = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{(6)^2+(-2)^2+(-4)^2}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14}\)
PR = \(\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{(9)^2+(-3)^2+(-6)^2}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}\)
Here, PQ + QR = \(\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}\) = PR
Hence, points P(-2, 3, 5), Q(1, 2, 3) & R(7, 0, -1) are collinear.
Question 2.
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 points (0, 7, -10), (1, 6, -6) & (4, 9, -6) are denoted by A, B & C respectively.
AB = \(\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}=\sqrt{(1)^2+(-1)^2+(4)^2}=\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2}\)
BC = \(\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2}=\sqrt{(3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
CA = \(\sqrt{(0-4)^2+(7-9)^2+(-10+6)^2}=\sqrt{(-4)^2+(-2)^2+(-4)^2}=\sqrt{16+4+16}=\sqrt{36}\) = 6
Here, AB = BC ≠ CA.
Thus, the given points are the vertices of an isosceles triangle.
ii) Let (0, 7, 10), (-1, 6, 6) & (-4, 9, 6) are denoted by A, B & C respectively.
AB = \(\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}=\sqrt{(-1)^2+(-1)^2+(-4)^2}=\sqrt{1+1+16}=\sqrt{18}\)
BC = \(\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}=\sqrt{(-3)^2+(3)^2+(0)^2}=\sqrt{9+9}=\sqrt{18}\)
CA = \(\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}=\sqrt{(4)^2+(-2)^2+(4)^2}=\sqrt{16+4+16}=\sqrt{36}\)
Now AB2 + BC2 = 18 + 18 = 36 = CA2.
Therefore, by Pythagoras theorem, ABC is a right angled triangle. Hence, the given points are the vertices of a right angled triangle.
iii) Let (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are denoted by A, B, C and D respectively.
AB = \(\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}=\sqrt{4+16+16}=\sqrt{36}\) = 6
BC = \(\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2}=\sqrt{9+25+9}=\sqrt{43}\)
CD = \(\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2}=\sqrt{4+16+16}=\sqrt{36}\) = 6
DA = \(\sqrt{(-1-2)^2+(2+3)^2+(1-4)^2}=\sqrt{9+25+9}=\sqrt{43}\)
Here AB = CD = 6 & BC = DA = \(\sqrt{43}\)
Hence, the opposite sides of quadrilateral ABCD are equal therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.
Question 3.
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 points A(1, 2, 3) & B(3, 2, -1). According to Geometric condition PA = PB
PA2 = PB2
⇒ (x – 1)2 + (y – 2)2 + (z – 3)2 = (x – 3)2 + (y – 2)2 + (z + 1)2
⇒ x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1
⇒ -2x – 4y – 6z + 14 = -6x – 4y + 2z + 14
⇒ -2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.
Question 4.
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:
Let the coordinates of P be (x, y, z). The coordinates of points A & B are (4, 0, 0) & (-4, 0, 0) respectively. It is given that PA + PB = 10.
⇒ \(\sqrt{(x-4)^2+y^2+z^2}+\sqrt{(x+4)^2+y^2+z^2}\) = 10
⇒ \(\sqrt{(x-4)^2+y^2+z^2}\) = 10 – \(\sqrt{(x-4)^2+y^2+z^2}\)
Squaring on both sides, we obtain
⇒ (x – 4)2 + y2 + z2 = 100 – \(\sqrt{(x+4)^2+y^2+z^2}\) + (x + 4)2 + y2 + z2
⇒ x2 – 8x + 16 + y2 + z2 = 100 – \(\sqrt{\left(x^2+8 x+16+y^2+z^2\right.}\) + x2 + 8x + 16 + y2 + z2
⇒ 20 \(\sqrt{x^2+8 x+16+y^2+z^2}\) = 100 + 16x
⇒ 5 \(\sqrt{x^2+8 x+16+y^2+z^2}\) = 25 + 4x
Squaring on both sides again, we have
25(x2 + 8x + 16 + y2 + z2) = 625 + 16x2 + 200x
⇒ 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x
⇒ 9x2 + 25y2 + 25z2 – 225 = 0
Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0.
Question 5.
If a variable point which moves such that 3PA = 2PB if A = (-2, 2, 3) and B = (13, -3, 13). Prove that P satisfies the equation x2 + y2 + z2 + 28x – 12y + 10z – 247 = 0.
Solution:
Let P(x, y, z) be any point on the locus. Given points are A = (-2, 2, 3), B = (13, -3,’ 13) and given geometric condition is 3PA = 2PB
⇒ 9PA2 = 4PB2
⇒ 9[(x + 2)2 + (y – 2)2 + (z – 3)2] = 4[(x – 13)2 + (y + 3)2 + (z – 13)2]
⇒ 9[x2 + 4 + 4x + y2 + 4 – 4y + z2 + 9 – 6z] = 4[x2 + 169 – 26x + y2 + 9 + 6y + z2 + 169 – 26z]
⇒ 5x2 + 5y2 + 5z2 + 140x – 60y + 50z – 1235 = 0
⇒ x2 + y2 + z2 + 28x – 12y + 10z – 247 = 0.