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Exercise 10.2 · Q16

Q.Find the position vector of the mid point of the vector joining the points P(2,3,4)P(2, 3, 4) and Q(4,1,−2)Q(4, 1, -2).

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The midpoint of a segment is the average of the endpoints’ coordinates. For P(2,3,4)P(2,3,4) and Q(4,1,−2)Q(4,1,-2), the midpoint’s position vector is 3i^+2j^+k^3\hat{i} + 2\hat{j} + \hat{k}.

Why the midpoint formula works

When you have two points in space, the vector from the origin to the midpoint is simply the average of the two position vectors. Think of it this way: if you walk from PP to QQ, the midpoint is exactly halfway along that journey. So you start at OP⃗\vec{OP}, then add half of the vector PQ⃗\vec{PQ} (which is OQ⃗−OP⃗\vec{OQ} - \vec{OP}). That gives:

OM⃗=OP⃗+12(OQ⃗−OP⃗)=OP⃗+OQ⃗2\vec{OM} = \vec{OP} + \frac{1}{2}(\vec{OQ} - \vec{OP}) = \frac{\vec{OP} + \vec{OQ}}{2}

This is the Section Formula for the midpoint — a special case of the more general internal division formula where the ratio is 1:11:1.

Midpoint position vector: OM⃗=OP⃗+OQ⃗2\displaystyle \vec{OM} = \frac{\vec{OP} + \vec{OQ}}{2}

Step-by-step solution

  1. Write the position vectors

    For P(2,3,4)P(2, 3, 4): OP⃗=2i^+3j^+4k^\vec{OP} = 2\hat{i} + 3\hat{j} + 4\hat{k}

    For Q(4,1,−2)Q(4, 1, -2): OQ⃗=4i^+1j^−2k^\vec{OQ} = 4\hat{i} + 1\hat{j} - 2\hat{k}

  2. Add the vectors component-wise

OP⃗+OQ⃗=(2+4)i^+(3+1)j^+(4−2)k^=6i^+4j^+2k^\vec{OP} + \vec{OQ} = (2+4)\hat{i} + (3+1)\hat{j} + (4-2)\hat{k} = 6\hat{i} + 4\hat{j} + 2\hat{k}

  1. Divide by 2 …

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