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Question 2 Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Class X1 - Maths -Introduction to Three Dimensional Geometry Page 277

Answers

Answered by abhi178
8
Let Q divides the Line joining point P and R in the ratio K : 1
P------------------------Q----------------R

use the section formula,
Q ≡ [ (mx2 + nx1)/(m + n) , (my2 + ny1)/(m+n) , (mz2 + nz1)/(m + n) ]
where, (x1, y1, z1) and (x2, y2, z2) are points and m:n is the ratio.

now,
Q ≡ [ (k × 9 + 1 × 3 )/(k + 1) , (k × 8 + 1 × 2 )/(k + 1) , ( k × -10 + 1 × -4 )/(k + 1)

but Q ≡ ( 5, 4, - 6) on comparing
5 = ( 9k + 3)/( k + 1)
5(k + 1) = 9k + 3
-4k = 3 - 5 = -2
k = 1/2

similarly,
4 = (8k + 2)/(k +1)
4(k + 1) = 8k + 2
4k + 4 = 8k + 2
-4k = -2 => k = 1/2

again, for z -axis
-6 = ( -10k - 4)/(k + 1)
-6k - 6 = -10k - 4
4k = 2 => k = 1/2

hence, point Q divides PR internally in the ratio 1 : 2
Answered by sharanbalaji2752007
0

Answer:

k=1:2

Step-by-step explanation:

Let point Q (5, 4, -6) divide the line segment joining points P (3, 2, -4) and R (9, 8, -10) in the ratio k:1.

Therefore, by section formula,

[\frac{mx2+nx1}{m+n},\frac{my2+my1}{m+n},\frac{mz2+mz1}{m+n}]

=[\frac{k(9)+3}{k+1},\frac{k(8)+2}{k+1},\frac{k(-10)-4}{k+1}]

=[\frac{9k+3}{k+1},\frac{8k+2}{k+1},\frac{-10k-4}{k+1}]

= > \frac{9k+3}{k+1}=5

= > 9k+3=5k+5

= > 4k=2

= > k=\frac{2}{4}

= > k=\frac{1}{2}

Thus, point Q divides PR in the ratio 1:2.

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