Learn Introduction to Three-dimensional Geometry with Top Tutors

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,y2−y1,z2−z1⟩ =⟨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.

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.