If $\to a$ and $\to b$ are the position vectors of points A and B respectively, find the position vector of point C on AB produced such that $\to A C = 3 \to A B$

2 \to a - 3 \to b

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3 \to a - 2 \to b

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3 \to b - 2 \to a

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2 \to b - 3 \to a

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Solution

The correct option is C $3 \to b - 2 \to a$

It is given in the question that $\to A C = 3 \to A B$
$\to B C = \to A C - \to A B$
$\to B C = 3 \to A B - \to A B$ $(\to A C = 3 \to A B)$
$\to B C = 2 \to A B$
Now $\frac{A C}{B C} = \frac{| \to A C |}{| \to B C |} = \frac{| 3 \to A B |}{| 2 \to A B |} = \frac{3}{2} \frac{| \to A B |}{| \to A B |} = \frac{3}{2}$
So C divides AB in the ratio 3:2 externally.
And we know that position vector of point P dividing $\to A B$ externally in the ratio m:n is given by $\frac{m \to B - n \to A}{m - n}$
Therefore the position vector of C is given by,
$\to r = \frac{3 \to b - 2 \to a}{3 - 2} = 3 \to b - 2 \to a$ , which is the correct answer.

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If →a and →b are the position vectors of points A and B respectively, find the position vector of point C on AB produced such that →AC=3→AB

If $\to a$ and $\to b$ are the position vectors of points A and B respectively, find the position vector of point C on AB produced such that $\to A C = 3 \to A B$