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Introduction to Three-dimensional Geometry

Introduction to Three-dimensional Geometry relates to CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry

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Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Introduction to Three-dimensional Geometry

Nazia Khanum

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for finding online coaching and tuition services. Now, let's tackle your math problem. To find the equation of the set of points equidistant from two given points, (1, 2, 3) and (3, 2, -1), we'll... read more

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for finding online coaching and tuition services. Now, let's tackle your math problem.

To find the equation of the set of points equidistant from two given points, (1, 2, 3) and (3, 2, -1), we'll use the concept of midpoint and distance formula in three dimensions.

First, let's find the midpoint of the line segment joining the two points. The midpoint is the average of the coordinates of the two points.

Midpoint MM is given by:

M(x1+x22,y1+y22,z1+z22)M(2x1+x2,2y1+y2,2z1+z2)

Substituting the coordinates of the given points:

M(1+32,2+22,3+(−1)2)M(21+3,22+2,23+(−1))

M(42,42,22)M(24,24,22)

M(2,2,1)M(2,2,1)

Now, we'll use the distance formula to find the distance from the midpoint to one of the given points. Since the points are equidistant, this distance will be the same for both points.

Distance formula:

d=(x2−x1)2+(y2−y1)2+(z2−z1)2d=(x2−x1)2+(y2y1)2+(z2z1)2

Let's take one of the given points, say (1, 2, 3):

d=(2−1)2+(2−2)2+(1−3)2d=(2−1)2+(2−2)2+(1−3)2

d=12+02+(−2)2d=12+02+(−2)2

d=1+0+4d=1+0+4

d=5d=5

Now, we have the midpoint (2, 2, 1) and the distance from the midpoint to one of the given points (55

). Using this information, we can write the equation of the set of points equidistant from the given points.

The equation would be:

(x−2)2+(y−2)2+(z−1)2=5(x−2)2+(y−2)2+(z−1)2

=5

This is the equation of the set of points equidistant from the points (1, 2, 3) and (3, 2, -1).

 
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Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Introduction to Three-dimensional Geometry

Nazia Khanum

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your math problem. Given that points P (3, 2, –4), Q (5, 4, –6), and R (9, 8, –10) are collinear, it means that they... read more

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your math problem.

Given that points P (3, 2, –4), Q (5, 4, –6), and R (9, 8, –10) are collinear, it means that they lie on the same straight line. Therefore, we can use the concept of section formula to find the ratio in which Q divides PR.

The section formula states that if a point Q divides the line segment joining two points P and R in the ratio m:n, then the coordinates of Q are given by:

Q(mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)Q(m+nmx2+nx1,m+nmy2+ny1,m+nmz2+nz1)

Where P has coordinates (x1,y1,z1)(x1,y1,z1), Q has coordinates (x,y,z)(x,y,z), and R has coordinates (x2,y2,z2)(x2,y2,z2).

Now, let's plug in the given coordinates:

x1=3x1=3, y1=2y1=2, z1=−4z1=−4

x2=9x2=9, y2=8y2=8, z2=−10z2=−10

m:nm:n is the ratio in which Q divides PR.

We need to find the ratio m:nm:n.

Since the points are collinear, the vector PR will be proportional to the vector PQ.

PR=⟨x2−x1,y2−y1,z2−z1⟩PR=⟨x2−x1,y2y1,z2z1⟩ =⟨9−3,8−2,−10−(−4)⟩=⟨9−3,8−2,−10−(−4)⟩ =⟨6,6,−6⟩=⟨6,6,−6⟩

Similarly, PQ=⟨5−3,4−2,−6−(−4)⟩PQ=⟨5−3,4−2,−6−(−4)⟩ =⟨2,2,−2⟩=⟨2,2,−2⟩

Now, to find the ratio m:nm:n, we compare the corresponding components of the vectors PQ and PR:

m6=n2=m+n−66m=2n=−6m+n

From the first two ratios, we get m=3nm=3n.

Substituting this into the third ratio, we get:

3n+n−6=−6−63n+n=−6

4n=−6(−6)4n=−6(−6)

n=−364n=4−36

n=−9n=−9

m=3n=3(−9)=−27m=3n=3(−9)=−27

Thus, the ratio in which Q divides PR is m:n=−27:−9m:n=−27:−9.

As an experienced tutor, I hope this explanation helps you understand how to solve similar problems in the future. If you have any further questions or need clarification on any concept, feel free to ask! And remember, for more personalized assistance, you can always find me on UrbanPro, one of the best online coaching platforms available.

 
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Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Introduction to Three-dimensional Geometry

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro is indeed one of the best platforms for online coaching and tuition. Now, to prove that the given points: (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) form the vertices of a right-angled triangle, we need to... read more

As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro is indeed one of the best platforms for online coaching and tuition. Now, to prove that the given points: (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) form the vertices of a right-angled triangle, we need to verify if one of the angles formed by these points is a right angle.

Firstly, we calculate the vectors formed by these points:

  • Vector AB = (0 - (-1), 7 - 6, 10 - 6) = (1, 1, 4)
  • Vector AC = (0 - (-4), 7 - 9, 10 - 6) = (4, -2, 4)

Now, to check if angle BAC is a right angle, we'll calculate the dot product of vectors AB and AC. If the dot product equals 0, it indicates that the vectors are perpendicular, implying that the angle between them is 90 degrees (a right angle).

AB ⋅ AC = (1)(4) + (1)(-2) + (4)(4) = 4 - 2 + 16 = 18

Since the dot product is not zero, it implies that the angle between vectors AB and AC is not 90 degrees. Therefore, the triangle formed by the points (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) is not a right-angled triangle.

However, I'd be happy to guide you through any other mathematical concepts or problems you might have. Feel free to reach out for further assistance! And remember, UrbanPro is your reliable resource for quality online coaching and tuition.

 
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Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Introduction to Three-dimensional Geometry

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into solving the problem. To calculate the perpendicular distance of point P(6, 7, 8) from the XY-plane, we need to find the distance between the point... read more

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into solving the problem.

To calculate the perpendicular distance of point P(6, 7, 8) from the XY-plane, we need to find the distance between the point and the XY-plane.

The equation of the XY-plane is z = 0, as it lies parallel to the XY-plane and passes through the origin.

The formula to calculate the distance between a point (x₁, y₁, z₁) and a plane ax + by + cz + d = 0 is given by:

Distance = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)

In this case, since the equation of the XY-plane is z = 0, the coefficients a, b, and c are all zero, and d = 0.

So, the distance reduces to:

Distance = |06 + 07 + 1*8 + 0| / √(0² + 0² + 1²) = |8| / √1 = 8

Therefore, the perpendicular distance of point P(6, 7, 8) from the XY-plane is 8 units.

Hence, the correct option is (a) 8.

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Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Introduction to Three-dimensional Geometry

Nazia Khanum

As an experienced tutor registered on UrbanPro, I can confidently guide you through this problem. UrbanPro is an excellent platform for online coaching and tuition, providing top-notch resources and expert guidance. To find the lengths of the edges of the parallelepiped formed by planes drawn through... read more

As an experienced tutor registered on UrbanPro, I can confidently guide you through this problem. UrbanPro is an excellent platform for online coaching and tuition, providing top-notch resources and expert guidance.

To find the lengths of the edges of the parallelepiped formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, we need to calculate the differences in coordinates between these points.

Let's denote the points as A(2, 3, 5) and B(5, 9, 7). The differences in coordinates give us the lengths of the edges of the parallelepiped along the x, y, and z directions.

Length of the edges:

  • Along x-axis: |5 - 2| = 3 units
  • Along y-axis: |9 - 3| = 6 units
  • Along z-axis: |7 - 5| = 2 units

Now, to find the length of the diagonal of the parallelepiped, we can use the distance formula in three dimensions.

Let's denote the diagonal as d. Using the coordinates of points A and B:

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Substituting the coordinates:

d = √[(5 - 2)² + (9 - 3)² + (7 - 5)²] = √[3² + 6² + 2²] = √(9 + 36 + 4) = √49 = 7 units

So, the lengths of the edges of the parallelepiped are 3 units, 6 units, and 2 units along the x, y, and z directions respectively. The length of the diagonal of the parallelepiped is 7 units. If you need further clarification or assistance, feel free to ask!

 
 
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