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Standard X

Mathematics

Section Formula

Find the posi...

Question

Find the position vector of a point which divides the join of points with position vectors $(\to a - 2 \to b)$ and $(2 \to a + \to b)$ externally in the ratio of $2 : 1$

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Solution

The position vector of the required point is given by $\frac{m \to A - n \to B}{m - n}$ , where the point divides the line $A B$ externally in the ratio $m : n$
Thus, we get $\frac{2 (\to a - 2 \to b) - 1 (2 \to a + \to b)}{2 - 1}$
i.e. $\frac{2 \to a - 4 \to b - 2 \to a - \to b}{1}$
$= - 5 \to b$

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\to a + 2 \to b, 2 \to a + \to b

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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 areexternally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

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Find the position vector of a point which divides the join of points with position vectors (→a−2→b) and (2→a+→b) externally in the ratio of 2:1

The position vector of the required point is given by m→A−n→Bm−n, where the point divides the line AB externally in the ratio m:nThus, we get 2(→a−2→b)−1(2→a+→b)2−1i.e. 2→a−4→b−2→a−→b1=−5→b

Find the position vector of a point which divides the join of points with position vectors $(\to a - 2 \to b)$ and $(2 \to a + \to b)$ externally in the ratio of $2 : 1$

The position vector of the required point is given by $\frac{m \to A - n \to B}{m - n}$ , where the point divides the line $A B$ externally in the ratio $m : n$
Thus, we get $\frac{2 (\to a - 2 \to b) - 1 (2 \to a + \to b)}{2 - 1}$
i.e. $\frac{2 \to a - 4 \to b - 2 \to a - \to b}{1}$
$= - 5 \to b$