Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Three Dimensional Coordinates Solutions Exercise 5(a) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Three Dimensional Coordinates Solutions Exercise 5(a)

I.

Question 1.

Find the distance of P (3, -2,4) from the origin.

Solution:

OP = \(\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{9+4+16}\)

= √29 units

Question 2.

Find the distance between the points (3,4,¬2) are (1,0,7).

Solution:

II.

Question 1.

Find x if the distance between (5, -1,7) and (x, 5,1) is 9 units.

Solution:

P(5, -1, 7), Q(x, 5,1) are the given points

PQ = 9

(5 – x)² + 36 + 36 = 81

(5 -x)² = 81 – 72 = 9

5 – x = ± 3

5 – x = 3 or 5 – x = -3

x = 5 – 3 or x = 5 + 3

= 2 or 8

Question 2.

Show that the points (2,3,5), (-1,5, -1) and (4, -3, 2) form a right angled iso-sceles triangle.

Solution:

A(2, 3, 5) B(-1, 5, -1), C(4, -3, 2) are the given points

AB² = (2 + 1)² + (3 – 5)² + (5 + 1)²

= 9 + 4 + 36 = 49

BC² = (-1- 4)² + (5 + 3)² + (-1 – 2)²

= 25 + 64 + 9 = 98

CA² = (4 – 2)² + (-3 -3)² + (2 – 5)²

= 4 + 36 + 9 = 49

AB² = CA² ⇒ AB² + CA² = 49 + 49 = 98 = BC²

ABC is a right angled isosceles triangle.

Question 3.

Show that the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) form an equilateral triangle.

Solution:

A( 1, 2, 3), B(2, 3, 1) and C (3, 1, 2) are the given points

AB² = (1 – 2)² + (2 – 3)²+ (3 – 1)²

= 1 + 1+ 4 = 6

BC² = (2 – 3)² + (3 – 1)² + (1 – 2)²

= 1 + 4 + 1 = 6

CA² = (3 – 1)² + (1 – 2)² + (2 – 3)²

= 4 + 1 + 6

AB² = BC² = CA² ⇒ AB = BC = CA

∴ ABC is an equilateral triangle

Question 4.

P is 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:

A(-2, 2, 3) and B(13, -3, 13) are the given points. P(x, y, z) is any point on the locus.

Given condition is 3P A = 2P B

⇒ 9 PA² = 4PA²

9[(x + 2)² + (y – 2)² + (z – 3)²] = 4[(x – 13)²+(y + 3)² + (z – 13)²]

⇒ 9(x² + 4x +4 + y² – 4y + 4 + z² – 6z + 9) = 4(x² – 26x + 169 + y² + 6y + 9 + z² – 26z + 169)

⇒ 9x² + 9y² + 9z² + 36x – 36y – 54z +153 = 4x² + 4y² + 4z² – 104x + 24y – 104z + 1388

5x² + 5f + 5z² + 140x – 60y + 50z – 1235 =0

Dividing with 5 locus of P is

x² + y² + z² + 28x – 12y + 10z – 247 = 0

Question 5.

Show that the points (1, 2, 3) (7, 0, 1) and (-2, 3, 4) are collinear.

Solution:

A (1, 2, 3), B(7, 0, 1) C(-2, 3, 4) are the given points.

AB + AC = 2√11 + √11 = 3√11 = BC

A, B, C are collinear

Question 6.

Show that ABCD is a square where A, B, C, D are the points (0, 4, 1), (2, 3 -1) (4, 5, 0) and (2, 6, 2) respectively.

Solution:

A(0, 4, 1), B(2, 3, -1) C(4, 5, 0), D(2, 6, 2) are the given points.

AB² = (0 – 2)² + (4 – 3)² + (1 + 1)²

=4 + 1 + 4 = 9

BC² = (2 – 4)² + (3 – 5)² + (-1 – 0)²

= 4 + 4 + 1 = 9

CD² = (4- 2) ²+ (5 – 6)² + (0 – 2)²

= 4 + 1 – 4 = 9

DA² = (2- 0) ²+ (6 – 4)² + (2 – 1)²

= 4 + 4 + 1 = 9

AB² = BC² = CD² = DA²

⇒ AB = BC = CD = DA

AC² = (0 – 4)² + (4 – 5)² + (1 – 0)² = 16 + 1 + 1 = 18

BD² = (2 – 2)² + (3 – 6)² + (-1 -2)² = 9 + 9 = 18

AC² = BD² ⇒ AC = BD

AB² + BC² = 9 + 9 = 18 = AC²

⇒ ∠ABC = 90°

A, B, C, D are the vertices of a square.