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Let-a-i-j-and-b-2i-k-the-point-of-intersection-of-the-lines-r-a-b-a-and-r-b-a-b-is-




Question Number 54420 by gunawan last updated on 03/Feb/19
Let a=i+j and b=2i−k, the point of  intersection of the lines r×a=b×a  and r×b=a×b is
$$\mathrm{Let}\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=\mathrm{2}\boldsymbol{\mathrm{i}}−\boldsymbol{\mathrm{k}},\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:\boldsymbol{\mathrm{r}}×\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}×\boldsymbol{\mathrm{a}} \\ $$$$\mathrm{and}\:\boldsymbol{\mathrm{r}}×\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{a}}×\boldsymbol{\mathrm{b}}\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
r=ix+jy+kz  r×a=[i  j   k  ] =i(0−z)−j(0−z)+k(x−y)                [x  y   z  ]               [1    1    0]  b×a=[i    j   k   ]=i(0+1)−j(0+1)+k(2)                [2    0  −1]                [1     1     0]  so  (−z)=1   (−z)=1    x−y=2 [ z=−1]]  r×b=[i   j       k   ]=i(−y−0)−j(−x−2z)+k(0−2y)                 [x  y    z     ]                 [2   0     −1]  a×b=−(b×a)=−i+j−2k  so  −y=−1 [y=1]   x+2z=1   x=1−2(−1)=3    x−y=2  x=2+1=3   [x=3   y=1   z=−1]  r=3i+j−k  pls write question clearly....
$${r}={ix}+{jy}+{kz} \\ $$$$\boldsymbol{{r}}×\boldsymbol{{a}}=\left[{i}\:\:{j}\:\:\:{k}\:\:\right]\:={i}\left(\mathrm{0}−{z}\right)−{j}\left(\mathrm{0}−{z}\right)+{k}\left({x}−{y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{x}\:\:{y}\:\:\:{z}\:\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{0}\right] \\ $$$$\boldsymbol{{b}}×\boldsymbol{{a}}=\left[\boldsymbol{{i}}\:\:\:\:\boldsymbol{{j}}\:\:\:\boldsymbol{{k}}\:\:\:\right]=\boldsymbol{{i}}\left(\mathrm{0}+\mathrm{1}\right)−\boldsymbol{{j}}\left(\mathrm{0}+\mathrm{1}\right)+\boldsymbol{{k}}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{2}\:\:\:\:\mathrm{0}\:\:−\mathrm{1}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}\right] \\ $$$$\left.\boldsymbol{{so}}\:\:\left(−\boldsymbol{{z}}\right)=\mathrm{1}\:\:\:\left(−\boldsymbol{{z}}\right)=\mathrm{1}\:\:\:\:\boldsymbol{{x}}−\boldsymbol{{y}}=\mathrm{2}\:\left[\:\boldsymbol{{z}}=−\mathrm{1}\right]\right] \\ $$$$\boldsymbol{{r}}×\boldsymbol{{b}}=\left[\boldsymbol{{i}}\:\:\:\boldsymbol{{j}}\:\:\:\:\:\:\:\boldsymbol{{k}}\:\:\:\right]=\boldsymbol{{i}}\left(−\boldsymbol{{y}}−\mathrm{0}\right)−\boldsymbol{{j}}\left(−\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{z}}\right)+\boldsymbol{{k}}\left(\mathrm{0}−\mathrm{2}\boldsymbol{{y}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{{x}}\:\:\boldsymbol{{y}}\:\:\:\:\boldsymbol{{z}}\:\:\:\:\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{2}\:\:\:\mathrm{0}\:\:\:\:\:−\mathrm{1}\right] \\ $$$$\boldsymbol{{a}}×\boldsymbol{{b}}=−\left(\boldsymbol{{b}}×\boldsymbol{{a}}\right)=−\boldsymbol{{i}}+\boldsymbol{{j}}−\mathrm{2}\boldsymbol{{k}} \\ $$$$\boldsymbol{{so}}\:\:−\boldsymbol{{y}}=−\mathrm{1}\:\left[\boldsymbol{{y}}=\mathrm{1}\right]\: \\ $$$$\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{z}}=\mathrm{1}\:\:\:\boldsymbol{{x}}=\mathrm{1}−\mathrm{2}\left(−\mathrm{1}\right)=\mathrm{3} \\ $$$$ \\ $$$$\boldsymbol{{x}}−\boldsymbol{{y}}=\mathrm{2}\:\:\boldsymbol{{x}}=\mathrm{2}+\mathrm{1}=\mathrm{3}\:\:\:\left[\boldsymbol{{x}}=\mathrm{3}\:\:\:\boldsymbol{{y}}=\mathrm{1}\:\:\:\boldsymbol{{z}}=−\mathrm{1}\right] \\ $$$$\boldsymbol{{r}}=\mathrm{3}\boldsymbol{{i}}+\boldsymbol{{j}}−\boldsymbol{{k}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{write}}\:\boldsymbol{{question}}\:\boldsymbol{{clearly}}…. \\ $$$$ \\ $$$$ \\ $$

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