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Physics Question on Vector basics

The position vector of a point RR which divides the line joining P(6, 3,−2)P(6,\,3,-2) and Q(3,1,−4)Q(3,1,-4) Q(3,1, - 4) in the ratio 2:12:1 externally is

A

i^+3j^−2k^\hat{i}+3\hat{j}-2\hat{k}

B

3i^−k^3\hat{i}-\hat{k}

C

−j^−6k^-\hat{j}-6\hat{k}

D

2i^−j^2\hat{i}-\hat{j}

Answer

−j^−6k^-\hat{j}-6\hat{k}

Explanation

Solution

Let position vector of P and Q are Day
OP=6i^+3j^−2k^OP=6\hat{i}+3\hat{j}-2\hat{k} and OQ=3i^+j^−4k^OQ=3\hat{i}+\hat{j}-4\hat{k}
Now, by section formula
OR=[2×OQ−1×OP2−1]OR=\left[ \frac{2\times OQ-1\times OP}{2-1} \right]
=(2(3i^+j^−4k^)−(6i^+3j^−2k^)1)=\left( \frac{2(3\hat{i}+\hat{j}-4\hat{k})-(6\hat{i}+3\hat{j}-2\hat{k})}{1} \right)
=\left\\{ \frac{6\hat{i}+2\hat{j}-8\hat{k}-6\hat{i}-3\hat{j}+2\hat{k}}{1} \right\\}
=(−j^  − 6k^)=(-\hat{j}\,\,-\,6\hat{k})