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Find-the-coordinate-of-the-point-in-R-3-which-is-the-reflection-the-point-1-2-3-with-respect-to-plane-X-Y-Z-1-

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Question Number 17520 by 1kanika# last updated on 07/Jul/17
Find the coordinate of the point in RΛ3 which is the reflection the point (1,2,3) with respect to plane X+Y+Z=1 .
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{R}\Lambda\mathrm{3}\:\mathrm{which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflection}\:\mathrm{the}\:\mathrm{point} \\ $$$$\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{plane}\: \\ $$$$\mathrm{X}+\mathrm{Y}+\mathrm{Z}=\mathrm{1}\:. \\ $$
Commented by 1kanika# last updated on 07/Jul/17
Here RΛ3 means R×R×R
$$\mathrm{Here}\:\mathrm{R}\Lambda\mathrm{3}\:\mathrm{means}\:\mathrm{R}×\mathrm{R}×\mathrm{R} \\ $$
Answered by ajfour last updated on 07/Jul/17
let given point is r_0 ^� =i^� +2j^� +3k^� equation of plane: r^� .n^� =q if reflected point is r_1 ^� =r_0 ^� +2λn^� then foot of perpendicular is r_M ^� =r_0 ^� +λn^� this point is in the plane, so r_M ^� .n^� =q ⇒ (r_0 ^� +λn^� ).n^� =q λ=((q−r_0 ^� .n^� )/(n^� .n^� )) =((1−(1+2+3))/(1+1+1)) =−(5/3) r_1 ^� = r_0 ^� +2λn^� =(i^� +2j^� +3k^� )−((10)/3)(i^� +j^� +k^� ) = (−(7/3)i^� −(4/3)j^� −(1/3)k^� ) hence image of given point is (−(7/3),−(4/3),−(1/3)) .
$$\mathrm{let}\:\mathrm{given}\:\mathrm{point}\:\mathrm{is}\:\:\bar {\mathrm{r}}_{\mathrm{0}} =\hat {\mathrm{i}}+\mathrm{2}\hat {\mathrm{j}}+\mathrm{3}\hat {\mathrm{k}} \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{plane}:\:\:\:\:\:\bar {\mathrm{r}}.\bar {\mathrm{n}}=\mathrm{q} \\ $$$$\mathrm{if}\:\mathrm{reflected}\:\mathrm{point}\:\mathrm{is}\:\:\bar {\mathrm{r}}_{\mathrm{1}} =\bar {\mathrm{r}}_{\mathrm{0}} +\mathrm{2}\lambda\bar {\mathrm{n}} \\ $$$$\mathrm{then}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{perpendicular}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\bar {\mathrm{r}}_{\mathrm{M}} =\bar {\mathrm{r}}_{\mathrm{0}} +\lambda\bar {\mathrm{n}} \\ $$$$\:\:\:\mathrm{this}\:\mathrm{point}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane},\:\mathrm{so} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\bar {\mathrm{r}}_{\mathrm{M}} .\bar {\mathrm{n}}=\mathrm{q} \\ $$$$\Rightarrow\:\:\:\:\:\left(\bar {\mathrm{r}}_{\mathrm{0}} +\lambda\bar {\mathrm{n}}\right).\bar {\mathrm{n}}=\mathrm{q} \\ $$$$\:\:\lambda=\frac{\mathrm{q}−\bar {\mathrm{r}}_{\mathrm{0}} .\bar {\mathrm{n}}}{\bar {\mathrm{n}}.\bar {\mathrm{n}}}\:=\frac{\mathrm{1}−\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)}{\mathrm{1}+\mathrm{1}+\mathrm{1}}\:=−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\bar {\mathrm{r}}_{\mathrm{1}} =\:\bar {\mathrm{r}}_{\mathrm{0}} +\mathrm{2}\lambda\bar {\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\hat {\mathrm{i}}+\mathrm{2}\hat {\mathrm{j}}+\mathrm{3}\hat {\mathrm{k}}\right)−\frac{\mathrm{10}}{\mathrm{3}}\left(\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(−\frac{\mathrm{7}}{\mathrm{3}}\hat {\mathrm{i}}−\frac{\mathrm{4}}{\mathrm{3}}\hat {\mathrm{j}}−\frac{\mathrm{1}}{\mathrm{3}}\hat {\mathrm{k}}\right) \\ $$$$\mathrm{hence}\:\mathrm{image}\:\mathrm{of}\:\mathrm{given}\:\mathrm{point}\:\mathrm{is} \\ $$$$\:\:\:\:\:\left(−\frac{\mathrm{7}}{\mathrm{3}},−\frac{\mathrm{4}}{\mathrm{3}},−\frac{\mathrm{1}}{\mathrm{3}}\right)\:. \\ $$
Commented by 1kanika# last updated on 09/Jul/17
Ans. of the ques. is (−3,−2,0)
$$\mathrm{Ans}.\:\mathrm{of}\:\mathrm{the}\:\mathrm{ques}.\:\mathrm{is}\:\left(−\mathrm{3},−\mathrm{2},\mathrm{0}\right) \\ $$
Commented by 1kanika# last updated on 09/Jul/17
Ans. of the ques. is (−3,−2,0)
$$\mathrm{Ans}.\:\mathrm{of}\:\mathrm{the}\:\mathrm{ques}.\:\mathrm{is}\:\left(−\mathrm{3},−\mathrm{2},\mathrm{0}\right) \\ $$
Commented by 1kanika# last updated on 09/Jul/17
Ans. of the ques. is (−3,−2,0)
$$\mathrm{Ans}.\:\mathrm{of}\:\mathrm{the}\:\mathrm{ques}.\:\mathrm{is}\:\left(−\mathrm{3},−\mathrm{2},\mathrm{0}\right) \\ $$
Commented by mrW1 last updated on 07/Jul/17
nice work!
$$\mathrm{nice}\:\mathrm{work}! \\ $$
Commented by mrW1 last updated on 09/Jul/17
distance from (1,2,3) to plane x+y+z=1 is 5/(√3) distance from (−7/3,−4/3,−1/3) to plane x+y+z=1 is 5/(√3) distance from (−3,−2,0) to plane x+y+z=1 is 6/(√3) ⇒ans. (−7/3,−4/3,−1/3) is right! ⇒ans. (−3,−2,0) is wrong!
$$\mathrm{distance}\:\mathrm{from}\:\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:\mathrm{to}\:\mathrm{plane}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1}\:\mathrm{is}\:\mathrm{5}/\sqrt{\mathrm{3}} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\left(−\mathrm{7}/\mathrm{3},−\mathrm{4}/\mathrm{3},−\mathrm{1}/\mathrm{3}\right)\:\mathrm{to}\:\mathrm{plane}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1}\:\mathrm{is}\:\mathrm{5}/\sqrt{\mathrm{3}} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\left(−\mathrm{3},−\mathrm{2},\mathrm{0}\right)\:\mathrm{to}\:\mathrm{plane}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1}\:\mathrm{is}\:\mathrm{6}/\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{ans}.\:\left(−\mathrm{7}/\mathrm{3},−\mathrm{4}/\mathrm{3},−\mathrm{1}/\mathrm{3}\right)\:\mathrm{is}\:\mathrm{right}! \\ $$$$\Rightarrow\mathrm{ans}.\:\left(−\mathrm{3},−\mathrm{2},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{wrong}! \\ $$
Commented by 1kanika# last updated on 09/Jul/17
we find reflcetion of a point not image of that point.
$$\mathrm{we}\:\mathrm{find}\:\mathrm{reflcetion}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point}\:\mathrm{not}\: \\ $$$$\mathrm{image}\:\mathrm{of}\:\mathrm{that}\:\mathrm{point}. \\ $$

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