Introduction to Three-dimensional Geometry: Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. Determine the octants wherein the values are positioned: (5, 2, 3).

Ans. The octants the values (5, 2, 3) are positioned in is I.

Ques. Determine the octants wherein the values are positioned: (-5, 4, 3)

Ans. The octants the values (-5, 4, 3) are positioned in is II.

Ques. Considering a point located on the X-axis, determine its coordinate.

Ans. The point located on the X-axis will have coordinates at (a, 0, 0).

Ques. Establish the image of (5, 2, -7) on a given xy plane.

Ans. The image of the following values on the xy plane can be otherwise represented as (5, 2, 7).

Ques. Establish the image of the values (-2, 3, 4) on a given yz plane.

Ans. The image of the following values on the yz plane can be otherwise represented as (2, 3, 4).

Ques. Highlight the name of a plane that houses both the x-axis and y-axis together.

Ans. The name of the plane that houses both x-axis and y-axis together is the XY plane.

Read More: Rolle’s Theorem

Short Answer Questions [2 Marks Questions]

Ques. What should be the distance between the origin to the point (a, b, c)?

Ans. The distance between the origin to the point is \(\sqrt{a^2 + b^2 + c^2}\)

Ques. Assuming that A, B, C on a typical x, y, z axes are the feet of perpendiculars ranging from point P, determine the values of coordinates A, B and C in (3, 4, 2).

Ans. It is already clear that y, z = 0 on the x-axis; x, z = 0 on the y-axis and x,y = 0 on the z-axis. As per the given data, we can consider the results. Assuming A, B, C are the feet of perpendiculars from point P, it can be said that the coordinates of each will be A(3, 0, 0), B(0, 4, 0), C(0, 0, 2).

Ques. Assuming that A, B, C on a typical x, y, z axes are the feet of perpendiculars ranging from point P, determine the values of coordinates A, B and C in (4, -3, -5)

Ans. As per the given data, we can consider the results. Assuming A, B, C are the feet of perpendiculars from point P, it can be said that the coordinates of each will be A(4, 0, 0), B(0, -3, 0), C(0, 0, -5).

Ques. Determine how distant the points (2,0, 0) and (-3, 0, 0) are located from one another.

Ans. As per the given equation, the points to highlight are, A (2, 0, 0) and B (-3, 0, 0).

Therefore, considering the statement, it can be otherwise determined that, AB = |2 – (-3)| = 5.

Ques. Determine how far apart the points P (3, 4, 5) are located from the given yz plane.

Ans. As per the given equation, the points to note are P (3,4, 5).

Hence, it can be said that, distance of P from the yz plane = |x coordinate of P| = 3

Thus, the points are found located at a distance of 3.

Ques. What should be a string’s total length that can be stretched in a rectangular tank which has dimensions 10, 13 and 8 units?

Ans. As per the given question, the dimensions of the rectangular tank are

A = 10

B = 13 

And, C = 8.

Thus, with the dimensions provided, we can simply determine the required length by the formula = \(\sqrt{a^2 + b^2 + c^2}\)

= \(\sqrt{10^2 + 13^2 + 8^2}\)

= \(\sqrt{100 + 169 + 64}\)

= \(\sqrt{333}\)
Thus, the required string length can be given as √333.

Ques. With the given sets of values, determine the distance each has from one another: (2, 3, 5) and (4, 3, 1).

Ans. As we are already aware, the distance between the given points P (x₁, y₁, z₁) and P (x₂, y₂, z₂) can be illustrated by,
PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)

Therefore, the distance can be shown from the values (2, 3, 5) and (4, 3, 1) that, 

= \(\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}\)

= \(\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}\)

= \(\sqrt{4+16}\)

= \(\sqrt{20}\)

= \(2\sqrt{5}\)

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Long Answer Questions [3 Marks Questions]

Ques. Determine the distance between the set of values,

1. (-1, 3, -4) and (1, -3, 4)
2. (-3, 7, 2) and (2, 4, -1)

Ans. As we are already aware, the distance between the given points P (x₁, y₁, z₁) and P (x₂, y₂, z₂) can be illustrated by,

(1) For values (-1, 3, -4) and (1, -3, 4), the distance is,

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

= \(\sqrt{(2)^{2}+(-6)^{2}+(8)^{2}}\)

= \(\sqrt{4+36+64}\)

= \(\sqrt{104}\)

= \(2\sqrt{26}\)

(2) For values (-3, 7, 2) and (2, 4, -1), the distance is,

= \(\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}\)

= \(\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}\)

= \(\sqrt{25+9+9}\)

= \(\sqrt{43}\)

Ques. Determine the ratio at which Q divides PR in the collinear equation values as P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10).

Ans. Let’s first consider that Q divides PR in the following ratio λ:1.
Therefore, the coordinator of Q automatically becomes,

\(\left(\frac{9 \lambda+3}{\lambda+1}, \frac{8 \lambda+2}{\lambda+1}, \frac{-10 \lambda-4}{\lambda+1}\right)\)

However, as per the coordinates as given for Q are (5, 4, -6), hence,

\(\left(\frac{9 \lambda+3}{\lambda+1}=5, \frac{8 \lambda+2}{\lambda+1}=4, \frac{-10 \lambda-4}{\lambda+1}=-6\right)\)

Which is to say, these three equations offer, 

λ = ½ 

Henceforth, Q is supposed to divide PR in the ratio of \(\frac{1}{2}\) : 1 or simply, 1:2

Ques. Establish an equation for a group or set of points that are kept equidistant from (1, 2, 3) and (3, 2, –1).

Ans. Looking at the equation as given above, consider that P (x, y, z) is the allotted point equidistant from A(1, 2, 3) and B(3, 2, –1)

Therefore, it is clear that PA = PB

Now, after considering to square both sides, we obtain,

PA² = PB²

Thus, it can be said that,

⇒ (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

Hence, after simplifying both ends, we should get,

⇒ –2x –4y – 6z + 14 = –6x – 4y + 2z + 14

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

Thus, the equation as required for the point set is x – 2z = 0.

Ques. Obtain the locus of a point that is found equidistant from A(0, 2, 3) and B(2, -2, 1).

Ans. To determine the value, let’s first consider that P (x, y, z) be any point equidistant from A(0, 2, 3) and B(2, -2, 1). So to say, PA = PB.

Hence, 

PA² = PB²

\(\Rightarrow \sqrt{(x-0)^{2}+(y-2)^{2}+(z-3)^{2}}=\sqrt{(x-2)^{2}+(y+2)^{2}+(z-1)^{2}}\)

⇒ 4x – 8y – 4z + 4 = 0

Or, x – 2y – z + 1 = 0

Thus, the locus of the points are determined.

Ques. Prove that points A(0,1 , 2), B(2, -1, 3) and C(1, -3, 1) are the respective vertices found in an isosceles right-angled triangle.

Ans. As Per the given equation, we can claim that,
AB = \(\sqrt{(2-0)^{2}+(-1-1)^{2}+(+3-2)^{2}}\)

= \(\sqrt{4 + 4 + 1 }\)

= 3

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

= \(\sqrt{1 + 4 + 4}\)

= 3

And, CA = \(\sqrt{(1-0)^{2}+(-3-1)^{2}+(1-2)^{2}}\)

= \(\sqrt{1 + 16 + 1}\)

= \(3\sqrt{2}\)

Which clearly establishes the fact that AB = BC and AB² + BC² = AC²
Therefore, it is proved that the triangle ABC is actually an isosceles right-angled triangle.

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Very Long Answer Questions [5 Marks Questions]

Ques. Highlight the points in the allotted xy plane that is shown in an equidistant position from points A (2, 0, 3), B(0, 3, 2) and C(0, 0, 1).

Ans. We are already aware that the z-coordinate found on every point on any xy plane is zero. Hence, consider P(x, y, 0) a point in the allotted xy plane, in a way that PA = PB = PC.

Now, it can be said, PA = PB

⇒ PA² = PB²

Thus,

⇒ (x - 2)² + (y - 0)²  + (0 - 3)² = (x - 0)² + (y - 3)²  + (0 - 2)²

⇒ 4x - 6y = 0 or else, 2x - 3y = 0 . . . (i)

Which is to say, PB = PC 

⇒ PB²  = PC²

Now, it can be claimed that,

⇒ (x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 0)² + (0 - 1)²

Therefore, 

⇒ -6y + 12 = 0

⇒ y = 2 . . . (ii)

Now, if we place y = 2 in the equation of (i), we will acquire x = 3

Thus, the points as required will be (3, 2, 0)

Ques. With the set of points as shown, prove that those are the vertices of a right angled triangle. The points are (0, 7, 10), (–1, 6, 6) and (–4, 9, 6).

Ans. As per the given equation, it can be said that the given points are A = (0, 7, 10), B = (–1, 6, 6), and C = (–4, 9, 6). Now, to determine the distance between them, we have to,

For the value of AB,

AB = √ [(-1-0)² + (6-7)² +(6-10)²]

AB = √ [(-1)² + (-1)² +(-4)²]

AB = √(1+1+16)

AB = √18

Therefore, AB = 3√2 . . . (i)

For the value of BC,

BC = √ [(-4+1)² + (9-6)² +(6-6)²]

BC = √ [(-3)² + (3)² +(0)²]

BC = √(9 + 9)

BC = √18

Therefore, BC = 3√2 . . . (ii)

For the value of CA,

CA = √ [(0+4)² + (7-9)² +(10-6)²]

CA = √ [(4)² + (-2)² +(4)²]

CA = √(16+4+16)

CA = √36

CA = 6 . . . (iii)

Now, after the values are determined, we need to apply Pythagoras Theorem, so as to,

AC² = AB² + BC² . . . (iv)

Following the same, we need to replace (1), (2), and (3) in (4),

6² = ( 3√2)² + ( 3√2)²

36 = 18 + 18

36 = 36

Thus, it is evident that the given points abide by the condition of Pythagoras Theorem which makes them the vertices of a right angled triangle.

Ques. It is said that the mid-points of the sides of a triangle ΔABC are provided by the values (-2, 3, 5), (4, -1, 7) and (6, 5, 3). With the given data, determine the coordinates of A, B and C.

Ans. Let’s first assume that the coordinates of ΔABC are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃). Now, let D, E and F be the mid-points of given sides BC, CA and AB respectively.

Therefore, \(\frac{x_1+x_2}{2} = 6\)

∴ x₁ + x₂ = 12 . . . (i)

\(\frac{y_1+y_2}{2} = 5\)

∴ y₁ + y₂ = 10 . . . (ii)

\(\frac{z_1+z_2}{2} = 3\)

∴ z₁ + z₂ = 6 . . . (iii)

\(\frac{x_2+x_3}{2} = -2\)

∴ x₂ + x₃ = -4 . . . (iv)

\(\frac{y_2+y_3}{2} = 3\)

∴ y₂ + y₃ = 6 . . . (v)

\(\frac{z_2+z_3}{2} = 5\)

∴ z₂ + z₃  = 10 . . . (vi)

\(\frac{x_1+x_3}{2} = 4\)

∴ x₁ + x₃ = 8 . . . (vii)

\(\frac{y_1+y_3}{2} = -1\)

∴ y₁ + y₃ = -2 . . . (viii)

\(\frac{z_1+z_3}{2} = 7\)

∴ z₁ + z₃ = 14 . . . (ix)

Now, after determining the values, we will add the equations i, iv, vii

2 (x₁ + x₂ + x₃) = 12 - 4 + 8

x₁ + x₂ + x₃ = \(\frac{16}{2}=8\) . . . (x)

Now, similarly, y₁ + y₂ + y₃ = 7 . . . (xi)

z₁ + z₂ + z₃ = 15 . . . (xii)

Now, it should follow subtraction of equations (i), (iv), and (vii) from (x)

x₃  = -4,

x₁ = 12

x₂ = 0

Following that, we should now subtract equations (ii), (v) and (viii) from (xi)

y₃ = - 3, y₁ = 1, y₂ = 9

Similarly so, z₃ = 9, z₁ = 5, z₂ = 1

Therefore, the coordinates of A, B and C are A(12, 0, -4), B(1, 9, -3) and C(5, 1, 9).

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