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Given, p=3hati+2hatj+4hatk,a=hati+hatj,b=hatj+hatk,c=hati+hatk and P = xa + yb + zc, then x, y, z are respectively |
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Answer» `(3)/(2), (1)/(2), (5)/(2)` `implies3hati+2hatj+4hatk=x(hati+hatj)+y(hatj+HATK)+z(hati+hatk)` `implies 3hati+2hatj+4hatk=(x+z)hati+(x+y)hatj+(y+z)hatk` On comparing both SIDES, the coefficients of `hati, hatj, hatk`, we get `x+z=3"... (i)"` `x+y=2"... (ii)"` and `y+z=4"... (iii)"` On solving EQS. (i), (ii) and (iii), we get `x=(1)/(2), y=(3)/(2), z=(5)/(2)` |
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