Given the position vectors v_1 = 2i−2j and v_2 =2j a) show that the unit vector in the direction of v_1 −v_2 = (1/(√5))(i−2j) b) Write down the equation of the line that contains the position vectors v_1 and v_2 c) Find the cosine of the angle between v_1 and v


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Question Number 48958 by Rio Michael last updated on 30/Nov/18

$G i v e n t h e p o s i t i o n v e c t o r s v_{1} = 2 i - 2 j a n d v_{2} = 2 j$ $a) s h o w t h a t t h e u n i t v e c t o r i n t h e d i r e c t i o n o f v_{1} - v_{2} = \frac{1}{\sqrt{5}} (i - 2 j)$ $b) W r i t e d o w n t h e e q u a t i o n o f t h e l i n e t h a t c o n t a i n s$ $t h e p o s i t i o n v e c t o r s v_{1} a n d v_{2}$ $c) F i n d t h e c o s i n e o f t h e a n g l e b e t w e e n v_{1} a n d v_{2}$
	Answered by ajfour last updated on 30/Nov/18

	$a) {\bar{v}}_{1} - {\bar{v}}_{2} = (2 \hat{i} - 2 \hat{j}) - 2 \hat{j} = 2 \hat{i} - 4 \hat{j}$ $s o {\hat{v}}_{12}, {\hat{v}}_{21} = \pm \frac{({\bar{v}}_{1} - {\bar{v}}_{2})}{∣ {\bar{v}}_{1} - {\bar{v}}_{2} ∣} = \pm \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j})$ $b) \bar{r} = ({\bar{v}}_{1} o r {\bar{v}}_{2}) + λ ({\bar{v}}_{1} - {\bar{v}}_{2})$ $\Rightarrow \bar{r} = 2 \hat{j} + λ (2 \hat{i} - 4 \hat{j})$ $c) \cos θ = \frac{{\bar{v}}_{1} . {\bar{v}}_{2}}{∣ {\bar{v}}_{1} ∣∣ {\bar{v}}_{2} ∣} = \frac{- 4}{2 \sqrt{2} \times 2} = - \frac{1}{\sqrt{2}} .$
	Commented by Rio Michael last updated on 09/Dec/18

	$s i r i d o n o t u n d e r s t a n d h o w d i d i t b e c o m e \pm \frac{1}{\sqrt{5}} (i - 2 j) ?$ $p l s s i m p l i f y m o r e$
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