7 Most VSAQ’s of Three Dimensional Coordinates Chapter in Inter 1st Year Maths-1B (TS/AP)

2 Marks

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:

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