Given the position vectors of points A and B are and.
Let the position vectors of points C and D be and.
We have AC = 3AB.
From the above figure, observe AB = AC – BC
⇒ AC = 3 (AC – BC)
⇒ AC = 3AC – 3BC
⇒ 2AC = 3BC
∴ AC : BC = 3 : 2
So, C divides AB externally in the ratio 3:2.
Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, externally in the ratio m : n is
Here, m = 3 and n = 2
So, the position vector of C is
We also have BD = 2BA.
From the figure, observe BA = BD – AD
⇒ BD = 2 (BD – AD)
⇒ BD = 2BD – 2AD
⇒ BD = 2AD
∴ BD : AD = 2 : 1
So, D divides BA externally in the ratio 2:1.
We now use the same formula as earlier to find the position vector of D.
Here, m = 2 and n = 1
Thus, the position vector of point C is and the position vector of point D is .
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