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Question Number 119588 by zakirullah last updated on 25/Oct/20

Answered by bemath last updated on 25/Oct/20

(i) let c^→  is unit vector such that  orthogonal to a^→  & b^→    so gives c^→  = ± ((∣c^→  ∣(a^→ ×b^→ ))/(∣a^→ ×b^→  ∣))  (•) a^→ ×b^→  =  determinant (((1   −2      3)),((2      1    −1   )))                      = (−1, 7, 5)  hence c^→  = ± (1/( (√(1+49+25)))) (−i+7j+5k)                     = ± (1/( 5(√3))) (−i+7j+5k )

$$\left({i}\right)\:{let}\:\overset{\rightarrow} {{c}}\:{is}\:{unit}\:{vector}\:{such}\:{that} \\ $$$${orthogonal}\:{to}\:\overset{\rightarrow} {{a}}\:\&\:\overset{\rightarrow} {{b}} \\ $$$$\:{so}\:{gives}\:\overset{\rightarrow} {{c}}\:=\:\pm\:\frac{\mid\overset{\rightarrow} {{c}}\:\mid\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right)}{\mid\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\:\mid} \\ $$$$\left(\bullet\right)\:\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\:=\:\begin{vmatrix}{\mathrm{1}\:\:\:−\mathrm{2}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{2}\:\:\:\:\:\:\mathrm{1}\:\:\:\:−\mathrm{1}\:\:\:}\end{vmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(−\mathrm{1},\:\mathrm{7},\:\mathrm{5}\right) \\ $$$${hence}\:\overset{\rightarrow} {{c}}\:=\:\pm\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{49}+\mathrm{25}}}\:\left(−{i}+\mathrm{7}{j}+\mathrm{5}{k}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\pm\:\frac{\mathrm{1}}{\:\mathrm{5}\sqrt{\mathrm{3}}}\:\left(−{i}+\mathrm{7}{j}+\mathrm{5}{k}\:\right) \\ $$$$ \\ $$

Commented by zakirullah last updated on 25/Oct/20

thanks alot

$$\boldsymbol{{thanks}}\:\boldsymbol{{alot}} \\ $$

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