Question Number 54419 by gunawan last updated on 03/Feb/19
$$\mathrm{If}\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}},\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{i}}−\boldsymbol{\mathrm{j}}+\boldsymbol{\mathrm{k}}\:\mathrm{and}\:\boldsymbol{\mathrm{c}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{unit} \\ $$$$\mathrm{vector}\:\bot\:\mathrm{to}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\mathrm{coplanar} \\ $$$$\mathrm{with}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}},\:\mathrm{then}\:\mathrm{a}\:\mathrm{unit}\:\mathrm{vector}\:\boldsymbol{\mathrm{d}}\:\bot\:\mathrm{to} \\ $$$$\mathrm{both}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{c}}\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
![c=xa+yb c=x(i+j−k)+y(i−j+k) c=i(x+y)+j(x−y)+k(−x+y) c.a=0 [i(x+y)+j(x−y)+k(−x+y)].[i+j−k]=0 (x+y)+(x−y)+(−x+y)(−1)=0 x+y+x−y+x−y=0 3x−y=0.....(1)eqn c=i(x+y)+j(x−y)+k(−x+y) c=i(x+3x)+j(x−3x)+k(−x+3x) c=i(4x)+j(−2x)+k(2x) given (√(16x^2 +4x^2 +4x^2 )) =1 (√(24x^2 )) =1 x=±(1/(√(24))) considering +ve value only...x=(1/(√(24))) y=(3/(√(24))) c=i(x+y)+j(x−y)+k(−x+y) c=i((4/(√(24))))+j(((−2)/(√(24))))+k((2/(√(24)))) considering −ve sign... c=i(((−4)/(√(24))))+j((2/(√(24))))+k(((−2)/(√(24)))) d=((a×c)/(∣a×c∣)) now a×c [i j k ] [1 1 −1 ] [(4/(√(24))) ((−2)/(√(24))) (2/(√(24)))] now you solve the remaining...](Q54437.png)
$${c}={x}\boldsymbol{{a}}+{y}\boldsymbol{{b}} \\ $$$$\boldsymbol{{c}}={x}\left({i}+{j}−{k}\right)+{y}\left({i}−{j}+{k}\right) \\ $$$$\boldsymbol{{c}}={i}\left({x}+{y}\right)+{j}\left({x}−{y}\right)+{k}\left(−{x}+{y}\right) \\ $$$$\boldsymbol{{c}}.\boldsymbol{{a}}=\mathrm{0} \\ $$$$\left[{i}\left({x}+{y}\right)+{j}\left({x}−{y}\right)+{k}\left(−{x}+{y}\right)\right].\left[{i}+{j}−{k}\right]=\mathrm{0} \\ $$$$\left({x}+{y}\right)+\left({x}−{y}\right)+\left(−{x}+{y}\right)\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}+{y}+{x}−{y}+{x}−{y}=\mathrm{0} \\ $$$$\mathrm{3}{x}−{y}=\mathrm{0}.....\left(\mathrm{1}\right){eqn} \\ $$$$\boldsymbol{{c}}={i}\left({x}+{y}\right)+{j}\left({x}−{y}\right)+{k}\left(−{x}+{y}\right) \\ $$$$\boldsymbol{{c}}={i}\left({x}+\mathrm{3}{x}\right)+{j}\left({x}−\mathrm{3}{x}\right)+{k}\left(−{x}+\mathrm{3}{x}\right) \\ $$$$\boldsymbol{{c}}={i}\left(\mathrm{4}{x}\right)+{j}\left(−\mathrm{2}{x}\right)+{k}\left(\mathrm{2}{x}\right) \\ $$$${given}\:\sqrt{\mathrm{16}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} }\:=\mathrm{1} \\ $$$$\sqrt{\mathrm{24}{x}^{\mathrm{2}} }\:=\mathrm{1}\:\:\:{x}=\pm\frac{\mathrm{1}}{\sqrt{\mathrm{24}}} \\ $$$${considering}\:+{ve}\:{value}\:{only}...{x}=\frac{\mathrm{1}}{\sqrt{\mathrm{24}}}\:\:\:{y}=\frac{\mathrm{3}}{\sqrt{\mathrm{24}}} \\ $$$$\boldsymbol{{c}}=\boldsymbol{{i}}\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)+\boldsymbol{{j}}\left(\boldsymbol{{x}}−\boldsymbol{{y}}\right)+\boldsymbol{{k}}\left(−\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$$\boldsymbol{{c}}=\boldsymbol{{i}}\left(\frac{\mathrm{4}}{\sqrt{\mathrm{24}}}\right)+\boldsymbol{{j}}\left(\frac{−\mathrm{2}}{\sqrt{\mathrm{24}}}\right)+\boldsymbol{{k}}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{24}}}\right) \\ $$$$\boldsymbol{{considering}}\:−\boldsymbol{{ve}}\:\boldsymbol{{sign}}... \\ $$$$\boldsymbol{{c}}=\boldsymbol{{i}}\left(\frac{−\mathrm{4}}{\sqrt{\mathrm{24}}}\right)+\boldsymbol{{j}}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{24}}}\right)+\boldsymbol{{k}}\left(\frac{−\mathrm{2}}{\sqrt{\mathrm{24}}}\right) \\ $$$$ \\ $$$$\boldsymbol{{d}}=\frac{\boldsymbol{{a}}×\boldsymbol{{c}}}{\mid\boldsymbol{{a}}×\boldsymbol{{c}}\mid}\:\:{now}\:\boldsymbol{{a}}×\boldsymbol{{c}} \\ $$$$\:\:\left[\boldsymbol{{i}}\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{j}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{k}}\:\:\:\right] \\ $$$$\:\:\:\left[\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\right] \\ $$$$\:\:\:\left[\frac{\mathrm{4}}{\sqrt{\mathrm{24}}}\:\:\:\:\:\frac{−\mathrm{2}}{\sqrt{\mathrm{24}}}\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{24}}}\right] \\ $$$${now}\:{you}\:{solve}\:{the}\:{remaining}... \\ $$