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Question: Let A and B be points with position vectors **a** and **b** with respect to the origin O. If the poi...

Let A and B be points with position vectors a and b with respect to the origin O. If the point C on OA is such that 2AC=CO,CD2AC = CO,CD is parallel to OB and ∣CD→∣=3∣OB→∣,|\overset{\rightarrow}{CD}| = 3|\overset{\rightarrow}{OB}|,

then AD→\overset{\rightarrow}{AD} is equal to

A

3b−a23\mathbf{b} - \frac{\mathbf{a}}{2}

B

3b+a23\mathbf{b} + \frac{\mathbf{a}}{2}

C

3b−a33\mathbf{b} - \frac{\mathbf{a}}{3}

D

3b+a33\mathbf{b} + \frac{\mathbf{a}}{3}

Answer

3b−a33\mathbf{b} - \frac{\mathbf{a}}{3}

Explanation

Solution

Since OA→=a,\overset{\rightarrow}{OA} = \mathbf{a}, OB→=b\overset{\rightarrow}{OB} = \mathbf{b} and 2AC=CO2AC = CO

By section formula OC→=23a.\overset{\rightarrow}{OC} = \frac{2}{3}\mathbf{a}.

Therefore, ∣CD→∣=3∣OB→∣⇒CD→=3b|\overset{\rightarrow}{CD}| = 3|\overset{\rightarrow}{OB}| \Rightarrow \overset{\rightarrow}{CD} = 3\mathbf{b}

⇒OD→=OC→+CD→=23a+3b\Rightarrow \overset{\rightarrow}{OD} = \overset{\rightarrow}{OC} + \overset{\rightarrow}{CD} = \frac{2}{3}\mathbf{a} + 3\mathbf{b}

Hence, AD→=OD→−OA→=23a+3b−a=3b−13a.\overset{\rightarrow}{AD} = \overset{\rightarrow}{OD} - \overset{\rightarrow}{OA} = \frac{2}{3}\mathbf{a} + 3\mathbf{b} - \mathbf{a} = 3\mathbf{b} - \frac{1}{3}\mathbf{a}.