TS Inter 1st Year Maths 1B Three-Dimensional Coordinates Important Questions

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Three-Dimensional Coordinates Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B Three-Dimensional Coordinates Important Questions

Question 1.
Find the distance between the points (3, 4, -2) and (1, 0, 7). [May ’00]
Solution:
Let A = (3, 4, -2), B = (1, 0, 7) are the given points.
Now,

Question 2.
Show that the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) form an equilateral triangle. [Mar. ’18 (AP); (B.P.)]
Solution:
Let A = (1, 2, 3), B = (2, 3, 1), C = (3, 1, 2) are the given points.
Now,


∴ Given points form an equilateral triangle.

Question 3.
Show that the points (1, 2, 3), (7, 0, 1) and (-2, 3, 4) are collinear. [Mar. ’16 (TS); Mar. ’13, May ’19]
Solution:
Let A = (1, 2, 3), B = (7, 0, 1), C = (-2, 3, 4) are the given points.
Now,

AB + CA = 2√11 + √11 = 3√11 = BC
∴ A, B, C are collinear.

Question 4.
Find the ratio in which yz-plane divides the line joining A(2, 4, 5) and B(3, 5, -4). Also, find the point of intersection. [May ’10]
Solution:
A(2, 4, 5), B(3, 5, -4) are the given points.
The ratio in which the yz-plane divides
\(\overline{\mathrm{AB}}\) = -x₁ : x₂ = -2 : 3
The point divides \(\overline{\mathrm{AB}}\) in the ratio -2 : 3,
then co-ordinates of a point = \(\left[\frac{\mathrm{mx}_2+\mathrm{nx_{1 }}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{my_{2 }}+\mathrm{ny_{1 }}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{mz_{2 }}+\mathrm{nz_{1 }}}{\mathrm{m}+\mathrm{n}}\right]\)

∴ The point of intersection = (0, 2, 23).

Question 5.
Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, -1), (3, 6, -1) and (4, 5, 1). [Mar. ’17 (TS), ’11; May ’03]
Solution:
A(2, 4, -1), B(3, 6, -1), C(4, 5, 1) are the given three vertices.
Let the fourth vertex be D(x, y, z)

3 + x = 6; 6 + y = 9; -1 + z = 0
x = 6 – 3; y = 9 – 6; z = 1
∴ x = 3, y = 3, z = 1
D = (3, 3, 1)
∴ Fourth Vertex D = (3, 3, 1)

Question 6.
Find the coordinates of the vertex ‘c’ of ∆ABC if its centroid is the origin and the vertices A, B are (1, 1, 1) and (-2, 4, 1) respectively. [Mar. ’16 (AP); ’15 (TS); May ’13, ’06]
Solution:
Let A(1, 1, 1), B(-2, 4, 1) are the given points.
Given that, Centroid G = (0, 0, 0)
Let third vertex, C = (x, y, z)
Now, the Centroid of ∆ABC is,

∴ The third vertex C = (1, -5, -2).

Question 7.
If (3, 2, -1), (4, 1, 1) and (6, 2, 5) are three vertices and (4, 2, 2) are the centroid of a tetrahedron, find the fourth vertex. [Mar. ’17, ’15 (AP), ’14, ’13 (old), ’09; May ’15 (AP), ’13, ’11, ’05]
Solution:
A(3, 2, -1), B(4, 1, 1), C(6, 2, 5) are the given points.
Given that, Centroid G = (4, 2, 2)
Let the fourth vertex, D = (x, y, z)
Now, the Centroid of tetrahedron ABCD is,


∴ The fourth vertex D = (3, 3, 3).

Question 8.
Show that the points A(3, 2, -4), B(5, 4, -6) and C(9, 8, -10) are collinear and find the ratio in which B divides \(\overline{\mathbf{A C}}\). [Mar. ’04]
Solution:
A = (3, 2, -4), B = (5, 4, -6), C = (9, 8, -10) are the given points.
Now,

Now, AB + BC = √12 + 2√12 = 3√12 = CA
∴ A, B, C are collinear.

The ratio in which ‘B’ divides \(\overline{\mathbf{A C}}\) = x₁ – x : x – x₂
= 3 – 5 : 5 – 9
= -2 : -4
= 2 : 4
= 1 : 2

Question 9.
If A(4, 8, 12), B(2, 4, 6), C(3, 5, 4) and D(5, 8, 5) are four points, show that the lines \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{CD}}\) intersect. [May ’04]
Solution:
A(4, 8, 12), B(2, 4, 6), C(3, 5, 4) and D(5, 8, 5) are the given points.
The equation of the line passing through A, B is

x – 4 = -2t; y – 8 = -4t; z – 12 = -6t
x = -2t + 4; y = -4t + 8; z = -6t + 12
∴ (x, y, z) = (-2t + 4, -4t + 8, -6t + 12) ……(1)
The equation of the line passing through C, D is

x – 3 = 2s; y – 5 = 3s; z – 4 = s
x = 2s + 3; y = 3s + 5; z = s + 4
∴ (x, y, z) = (2s + 3, 3s + 5, s + 4) ………(2)
From (1) &(2)
-2t + 4 = 2s + 3 ⇒ 2t + 2s – 1 = 0 ………(3)
-4t + 8 = 3s + 5 ⇒ 4t + 3s – 3 = 0 ………(4)
-6t + 12 = s + 4 ⇒ 6t + s – 8 = 0 ……….(5)
Solving (3) & (4)


Substitute the values of (t, s) in equation (5)
6(\(\frac{3}{2}\)) + (-1) – 8 = 0
9 – 9 = 0
0 = 0
∴ Lines \(\overline{\mathrm{AB}} \& \overline{\mathrm{CD}}\) are intersecting lines.
Substitute, t = \(\frac{3}{2}\) in equation (1) or s = -1 in equation (2)
∴ Point of intersection = [-2(\(\frac{3}{2}\)) + 4, -4(\(\frac{3}{2}\)) + 8, -6(\(\frac{3}{2}\)) + 12]
= [-3 + 4, -6 + 8, -9 + 12]
= (1, 2, 3)

Question 10.
Find the point which divides the line joining the points A(2, -3, 1), B(3, 4, -5) in the ratio 1 : 3.
Solution:
A(2, -3, 1), B(3, 4, -5) are the given points.
Let C(x, y, z) be the point which divides the line joining the points A(2, -3, 1), B(3, 4, -5) in the ratio 1 : 3.

Question 11.
Find the centroid of the triangle whose vertices are (5, 4, 6), (1, -1, 3) and (4, 3, 2).
Solution:
A(5, 4, 6), B(1, -1, 3), C(4, 3, 2) are the given vertices.
The centroid of the triangle ABC is

Question 12.
Find the centroid of the tetrahedron whose vertices are (2, 3, -4), (-3, 3, -2), (-1, 4, 2), (3, 5, 1).
Solution:
A(2, 3, -4), B(-3, 3, -2), C(-1, 4, 2) and D(3, 5, 1) are the given points.
The centroid of the tetrahedron is

Some More Maths 1B Three-Dimensional Coordinates Important Questions

Question 13.
Find the distance of P(3, -2, 4) from the origin.
Solution:
Let O(0, 0, 0), P(3, -2, 4) are the given points.
∴ Distance from origin to P(3, -2, 4) is

Question 14.
Show that the points A(3, -2, 4), B(1, 1, 1) and C(-1, 4, -2) are collinear.
Solution:
Let A = (3, -2, 4), B = (1, 1, 1), C = (-1, 4, -2) are the given points.
Now,


AB + BC = √22 + √22 = 2√22 = AC
∴ A, B, C are collinear.

Question 15.
Show that the points (5, 4, 2), (6, 2, -1) and (8, -2, -7) are collinear. [May ’07]
Solution:
Let A = (5, 4, 2), B = (6, 2, -1), C = (8, -2, -7) are the given points.

Now, AB + BC = √14 + 2√14 = 3√14 = CA
∴ A, B, C are collinear.

Question 16.
Find the ratio in which the xz-plane divides the line joining A(-2, 3, 4) and B(1, 2, 3).
Solution:
A(-2, 3, 4), B(1, 2, 3) are the given points.
xz-plane divides \(\overline{\mathrm{AB}}\) in the ratio = -y₁ : y₂
= -3 : 2 (3 : 2 externally)

Question 17.
Find the point of intersection of the lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) where A = (7, -6, 1), B = (17, -18, -3), C = (1, 4, -5) and D = (3, -4, 11).
Solution:
A(7, -6, 1), B(17, -18, -3), C(1, 4, -5) and D(3, -4, 11) are the given points.
The equation of the line passing through A, B is

x – 7 = 10t; y + 6 = -12t; z – 1 = 4t
x = 10t + 7; y = -12t – 6; z = -4t + 1
∴ (x, y, z) = (10t + 7, -12t – 6, -4t + 1) ……..(1)
The equation of the line passing through C, D is

x – 1 = 2s; y – 4 = -8s; z + 5 = 16s
x = 2s + 1; y = -8s + 4; z = 16s – 5
∴ (x, y, z) = (2s + 1, -8s + 4, 16s – 5) ……….(2)
From (1) & (2)
10t + 7 = 2s + 1
⇒ 10t – 2s + 6 = 0
⇒ 5t – s + 3 = 0 ……..(3)
-12t – 6 = -8s + 4
⇒ 12t – 8s + 10 = 0
⇒ 6t – 4s + 5 = 0 ……..(4)
-4t + 1 = 16s – 5
⇒ 4t + 16s – 6 = 0
⇒ 2t + 8s – 3 = 0 ………(5)
Solving (3) & (4)

Substitute the values of t, s in equation (5)
\(2\left(\frac{-1}{2}\right)+8\left(\frac{1}{2}\right)-3=0\)
-1 + 4 – 3 = 0
⇒ 4 – 4 = 0
⇒ 0 = 0
∴ Lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are intersecting lines.
Substitute t = \(\frac{-1}{2}\) in equation (1) & s = \(\frac{1}{2}\) in equation (2)
∴ Point of intersection = \(\left[10\left(\frac{-1}{2}\right)+7,-12\left(\frac{-1}{2}\right)-6,-4\left(\frac{-1}{2}\right)+1\right]\)
= [-5 + 7, 6 – 6, 2 + 1]
= (2, 0, 3)

Question 18.
Show that the points A(-4, 9, 6), B(-1, 6, 6) and C(0, 7, 10) form a right angled isoscele.
Solution:
Let A = (-4, 9, 6), B = (-1, 6, 6), C = (0, 7, 10) are the given points triangle.
Now,

AB = CA then triangle ABC is isosceles.
AB² + BC² = (√18)² + (√18)²
= 18 + 18
= (√36)²
= AC²
∴ AB² + BC² = AC² then ΔABC is right-angled.
∴ ΔABC is a right-angled isosceles triangle.

Question 19.
Find ‘x’ if the distance between (5, -1, 7) and (x, 5, 1) is 9 units. [Mar. ’19 (AP)]
Solution:
Let A = (5, -1, 7), B = (x, 5, 1) are the given points
Now, Given that, AB = 9

Squaring on both sides
x² – 10x + 97 = 81
⇒ x² – 10x + 16 = 0
⇒ x² – 8x – 2x + 16 = 0
⇒ x(x – 8) – 2(x – 8) = 0
⇒ (x – 8) (x – 2) = 0
⇒ x = 8 or 2

Question 20.
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), and D = (2, 6, 2) are the given points.
Now,


∴ Given points from a square.
∴ AB = BC = CD = DA & AC = BD

Question 21.
If (x₁, y₁, z₁) and (x₂, y₂, z₂) are two vertices and (α, β, γ) is the centroid of a triangle, find the third vertex of the triangle.
Solution:
Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be the two vertices and C(x, y, z) be the third vertex.
Given G = (α, β, γ) we have
\(\frac{x+x_1+x_2}{3}\) = α, \(\frac{y+y_1+y_2}{3}\) = β, \(\frac{z+z_1+z_2}{3}\) = γ
∴ x = 3α – x₁ – x₂, y = 3β – y₁ – y₂, z = 3γ – z₁ – z₂
∴ The third vertex = (3α – x₁ – x₂, 3β – y₁ – y₂, 3γ – z₁ – z₂)

Question 22.
If M(α, β, γ) is the midpoint of the line segment joining the points A(x₁, y₁, z₁) and B then find B.
Solution:
Let B = (x, y, z) then coordinates of midpoint = \(\left(\frac{x_1+x}{2}, \frac{y_1+y}{2}, \frac{z_1+z}{2}\right)\)
Given (α, β, γ) = \(\left(\frac{x_1+x}{2}, \frac{y_1+y}{2}, \frac{z_1+z}{2}\right)\)
∴ x = 2α – x₁, y = 2β – y₁, z = 2γ – z₁
∴ B = (2α – x₁, 2β – y₁, 2γ – z₁)

Question 23.
If H, G, S, and I respectively denote the orthocentre, centroid, circumcentre, and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1), and (3, 1, 2) then find H, G, S, I.
Solution:
Let A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) be three given points.

∴ AB = BC = CA, the triangle formed will be an equilateral triangle.
In this triangle, all the centres H, G, I and S coincide.
∴ Centroid of the triangle (G) = \(\left(\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}\right)\) = (2, 2, 2)
Hence G = H = S = I = (2, 2, 2)

Question 24.
Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0), and (0, 4, 0).
Solution:
If A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) are the vertices of a triangle
and a = BC, b = CA and c = AB are the sides of the triangle then the incentre of the triangle,

Question 25.
Find the distance between the midpoint of the line segment \(\overline{\mathrm{AB}}\) and the point (3, -1, 2) where A = (6, 3, -4) and B = (-2, -1, 2)?
Solution:
Let A = (6, 3, -4), B = (-2, -1, 2) are the given points.
Midpoint of \(\overline{\mathrm{AB}}\) is

Let D = (3, -1, 2) be given point
Now, the distance between C & D