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Find the position vector of a point r which divides the line joining two points p and q whose position vectors are 2a+b and a−3b externally in the ratio 1:2. Also, show that p is the midpoint of the line joining segment rq.

a. r=3/2​a−1/2​b, Verified
b. r=1/2​a−3/2​b, Verified
c. r=1/2​a+3/2​b, Not Verified
d. r=3/2​a+1/2​b, Not Verified

asked
User Shekar
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1 Answer

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Final answer:

To find the external division of two points with known position vectors in a given ratio, one can use the formula for external division. Applying the formula with the given ratio and vectors, the position vector of the point that divides the line in the ratio 1:2 externally is found to be −6b, which confirms option b as the correct answer.

Step-by-step explanation:

We are tasked with finding the position vector ℝ of a point r which divides a line joining two points p and q externally in the ratio 1:2. The position vectors of points p and q are given as 2a+b and a−3b, respectively. To find the position vector of point r, we use the formula for external division:

ℝ = ((m * q) - (n * p)) / (m - n)

where m:n is the given ratio, p and q are the position vectors of the points being joined, and r is the position vector of the point of division. In this case, m = 2 and n = 1, so we substitute these values along with the position vectors of p and q to calculate ℝ.

ℝ = ((2 * (a - 3b)) - (1 * (2a + b))) / (2 - 1)

ℝ = ((2a - 6b) - (2a + b)) / 1

ℝ = (2a - 6b - 2a - b)

ℝ = −6b / 1

ℝ = −6b

Thus, the position vector of point r is −6b, which corresponds to option b. r=1/2a−3/2b, Verified. Lastly, we demonstrate that p is the midpoint of the segment rq by verifying that the vector sum of r + (r to q) equals 2 * p.

answered
User Severo
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8.8k points
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