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Question Number 48958 by Rio Michael last updated on 30/Nov/18
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_2
$${Given}\:{the}\:{position}\:{vectors}\:{v}_{\mathrm{1}} =\:\mathrm{2}\boldsymbol{{i}}−\mathrm{2}\boldsymbol{{j}}\:{and}\:{v}_{\mathrm{2}} =\mathrm{2}\boldsymbol{{j}} \\ $$$$\left.{a}\right)\:{show}\:{that}\:{the}\:{unit}\:{vector}\:{in}\:{the}\:{direction}\:{of}\:{v}_{\mathrm{1}} −{v}_{\mathrm{2}} =\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left({i}−\mathrm{2}{j}\right) \\ $$$$\left.{b}\right)\:{Write}\:{down}\:{the}\:{equation}\:{of}\:{the}\:{line}\:{that}\:{contains} \\ $$$${the}\:{position}\:{vectors}\:{v}_{\mathrm{1}} \:{and}\:{v}_{\mathrm{2}} \\ $$$$\left.{c}\right)\:{Find}\:{the}\:{cosine}\:{of}\:{the}\:{angle}\:{between}\:{v}_{\mathrm{1}} \:{and}\:{v}_{\mathrm{2}} \\ $$
Answered by ajfour last updated on 30/Nov/18
a) v_1 ^� −v_2 ^� =(2i^� −2j^� )−2j^� = 2i^� −4j^� so v_(12) ^� ,v_(21) ^� =± (((v_1 ^� −v_2 ^� ))/(∣v_1 ^� −v_2 ^� ∣)) = ±(1/( (√5)))(i^� −2j^� ) b) r^� = (v_1 ^� or v_2 ^� )+λ(v_1 ^� −v_2 ^� ) ⇒ r^� = 2j^� +λ(2i^� −4j^� ) c) cos θ = ((v_1 ^� .v_2 ^� )/(∣v_1 ^� ∣∣v_2 ^� ∣)) = ((−4)/(2(√2)×2)) = −(1/( (√2))) .
$$\left.{a}\right)\:\bar {{v}}_{\mathrm{1}} −\bar {{v}}_{\mathrm{2}} =\left(\mathrm{2}\hat {{i}}−\mathrm{2}\hat {{j}}\right)−\mathrm{2}\hat {{j}}\:=\:\mathrm{2}\hat {{i}}−\mathrm{4}\hat {{j}} \\ $$$${so}\:\:\hat {{v}}_{\mathrm{12}} \:,\hat {{v}}_{\mathrm{21}} =\pm\:\frac{\left(\bar {{v}}_{\mathrm{1}} −\bar {{v}}_{\mathrm{2}} \right)}{\mid\bar {{v}}_{\mathrm{1}} −\bar {{v}}_{\mathrm{2}} \mid}\:=\:\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\hat {{i}}−\mathrm{2}\hat {{j}}\right) \\ $$$$\left.{b}\right)\:\:\:\:\bar {{r}}\:=\:\left(\bar {{v}}_{\mathrm{1}} \:{or}\:\bar {{v}}_{\mathrm{2}} \right)+\lambda\left(\bar {{v}}_{\mathrm{1}} −\bar {{v}}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\bar {{r}}\:=\:\mathrm{2}\hat {{j}}+\lambda\left(\mathrm{2}\hat {{i}}−\mathrm{4}\hat {{j}}\right) \\ $$$$\left.{c}\right)\:\:\mathrm{cos}\:\theta\:=\:\frac{\bar {{v}}_{\mathrm{1}} .\bar {{v}}_{\mathrm{2}} }{\mid\bar {{v}}_{\mathrm{1}} \mid\mid\bar {{v}}_{\mathrm{2}} \mid}\:=\:\frac{−\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}}×\mathrm{2}}\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:. \\ $$
Commented by Rio Michael last updated on 09/Dec/18
sir i do not understand how did it become ±(1/( (√5)))(i−2j)? pls simplify more
$${sir}\:{i}\:{do}\:{not}\:{understand}\:{how}\:{did}\:{it}\:{become}\:\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left({i}−\mathrm{2}{j}\right)? \\ $$$${pls}\:{simplify}\:{more} \\ $$

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