Three Dimensional Coordinates (VSAQs)

Maths-1B | 5. Three Dimensional Coordinates – VSAQs:
Welcome to VSAQs in Chapter 5: Three Dimensional Coordinates. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : 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

Given vertices A = (1,1,1) B = (-2,4,1) and the third vertex $$C = (x_3,y_3,z_3)$$ nd the centroid G = (0,0,0)

Calculating the centroid of $$\triangle ABC$$

$$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$$

$$\Rightarrow \left(\frac{1 – 2 + x_3}{3}, \frac{1 + 4 + y_3}{3}, \frac{1 + 1 + z_3}{3}\right) = (0, 0, 0)$$

$$\Rightarrow \left(\frac{-1 + x_3}{3}, \frac{5 + y_3}{3}, \frac{2 + z_3}{3}\right) = (0, 0, 0)$$

From this we derive:

$$\frac{-1 + x_3}{3} = 0$$

$$x_3 – 1 = 0$$

$$x_3 = 1$$

$$\frac{5 + y_3}{3} = 0$$

$$y_3 + 5 = 0$$

$$y_3 = -5$$

$$\frac{2 + z_3}{3} = 0$$

$$z_3 + 2 = 0$$

$$z_3 = -2$$

Hence, the third vertex C is determined to be (1, -5, -2)

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VSAQ-2 : If (3, 2, -1), (4, 1, 1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of a tetrahedron find the fourth vertex of that tetrahedron

Given vertices:

$$A = (3, 2, -1)$$

$$B = (4, 1, 1)$$

$$C = (6, 2, 5)$$

$$D = (x_4, y_4, z_4)$$

and given centroid $$G = (4, 2, 2)$$

To find the centroid of the tetrahedron ABCD

$$G = \left(\frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4}\right)$$

Substituting the given coordinates and the centroid:

$$\Rightarrow \left(\frac{3 + 4 + 6 + x_4}{4}, \frac{2 + 1 + 2 + y_4}{4}, \frac{-1 + 1 + 5 + z_4}{4}\right) = (4, 2, 2)$$

$$\Rightarrow \left(\frac{13 + x_4}{4}, \frac{5 + y_4}{4}, \frac{5 + z_4}{4}\right) = (4, 2, 2)$$

$$13 + x_4 = 16$$

$$x_4 = 16 – 13 = 3$$

$$5 + y_4 = 8$$

$$y_4 = 8 – 5 = 3$$

$$5 + z_4 = 8$$

$$z_4 = 8 – 5 = 3$$

Thus, the coordinates for vertex D are determined to be D = (3, 3, 3)

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VSAQ-3 : Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, -1), (3, 6, -1) and (4, 5, 1)

Given vertices:

$$A = (2, 4, -1)$$

$$B = (3, 6, -1)$$

$$C = (4, 5, 1)$$

$$D = (a, b, c)$$

In the parallelogram ABCD the midpoint of AC must equal the midpoint of BD

$$\Rightarrow \left(\frac{2 + 4}{2}, \frac{4 + 5}{2}, \frac{-1 + 1}{2}\right) = \left(\frac{3 + a}{2}, \frac{6 + b}{2}, \frac{-1 + c}{2}\right)$$

From this, we derive:

For a:

$$3 + \frac{a}{2} = 3$$

$$\frac{a}{2} = 0$$

$$a = 0$$

For b:

$$4.5 = \frac{6 + b}{2}$$

$$9 = 6 + b$$

$$b = 3$$

For c:

$$0 = \frac{-1 + c}{2}$$

$$0 = -1 + c$$

$$c = 1$$

Thus, the coordinates for vertex D are determined to be D = (3, 3, 1)

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VSAQ-4 : Find the ratio in which the XZ – plane divides line joining A (-2, 3, 4) & B (1, 2, 3) Also find the point of intersection

Given:

$$A(x_1, y_1, z_1) = (-2, 3, 4)$$

$$B(x_2, y_2, z_2) = (1, 2, 3)$$

The ratio in which the XZ plane divides AB is given by the y-coordinates of A and B

$$-y_1 : y_2 = -3 : 2$$

$$P = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}, \frac{m z_2 + n z_1}{m + n}\right)$$

Plugging in the values and ratio:

$$m = -3, n = 2$$

$$P = \left(\frac{-3 \cdot 1 + 2 \cdot (-2)}{-3 + 2}, \frac{-3 \cdot 2 + 2 \cdot 3}{-3 + 2}, \frac{-3 \cdot 3 + 2 \cdot 4}{-3 + 2}\right)$$

$$P = \left(\frac{-3 – 4}{-1}, \frac{-6 + 6}{-1}, \frac{-9 + 8}{-1}\right)$$

$$P = \left(\frac{-7}{-1}, \frac{0}{-1}, \frac{-1}{-1}\right)$$

$$P = (7, 0, 1)$$

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VSAQ-5 : 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

Given points:

$$A(x_1, y_1, z_1) = (2, 4, 5)$$

$$B(x_2, y_2, z_2) = (3, 5, -4)$$

The ratio that the yz-plane divides AB is given by $$-x_1:x_2 = -2:3$$

Using the section formula for internal division:

$$P = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}, \frac{m z_2 + n z_1}{m + n}\right)$$

Where m = -2 and n = 3 substituting into the formula:

$$P = \left(\frac{-2 \times 3 + 3 \times 2}{-2 + 3}, \frac{-2 \times 5 + 3 \times 4}{-2 + 3}, \frac{-2 \times (-4) + 3 \times 5}{-2 + 3}\right)$$

$$P = \left(\frac{-6 + 6}{1}, \frac{-10 + 12}{1}, \frac{8 + 15}{1}\right)$$

$$P = (0, 2, 23)$$

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VSAQ-6 : Show that the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) form an equilateral triangle

Given vertices:

$$A = (1, 2, 3)$$

$$B = (2, 3, 1)$$

$$C = (3, 1, 2)$$

Calculating the lengths of the sides AB BC and AC using the distance formula:

Distance AB

$$AB = \sqrt{(2 – 1)^2 + (3 – 2)^2 + (1 – 3)^2}$$

$$AB = \sqrt{1 + 1 + 4} = \sqrt{6}$$

Distance BC

$$BC = \sqrt{(3 – 2)^2 + (1 – 3)^2 + (2 – 1)^2}$$

$$BC = \sqrt{1 + 4 + 1} = \sqrt{6}$$

Distance AC

$$AC = \sqrt{(3 – 1)^2 + (1 – 2)^2 + (2 – 3)^2}$$

$$AC = \sqrt{4 + 1 + 1} = \sqrt{6}$$

$$AB = BC = AC$$

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

Given vertices:

$$A = (1, 2, 3)$$

$$B = (7, 0, 1)$$

$$C = (-2, 3, 4)$$

Distance AB

$$AB = \sqrt{(7 – 1)^2 + (0 – 2)^2 + (1 – 3)^2}$$

$$AB = \sqrt{36 + 4 + 4} = \sqrt{44} = 2\sqrt{11}$$

Distance BC

$$BC = \sqrt{(-2 – 7)^2 + (3 – 0)^2 + (4 – 1)^2}$$

$$BC = \sqrt{81 + 9 + 9} = \sqrt{99} = 3\sqrt{11}$$

Distance AC

$$AC = \sqrt{(-2 – 1)^2 + (3 – 2)^2 + (4 – 3)^2}$$

$$AC = \sqrt{9 + 1 + 1} = \sqrt{11}$$

Checking for Collinearity:

$$AB + AC = 2\sqrt{11} + \sqrt{11} = 3\sqrt{11} = BC$$