Given ` vecP=3 hati+2hatj+5hatk, veca=hati+hatj, vecb=hatj+hatk, vecc=hati+hatk and vecP=xveca+yvecb+zvecc`, then x,y,z are respectivelyA. `(3)/(2), (1)/(2), (5)/(2)`B. `(1)/(2),(3)/(2),(5)/(2)`C. `(5)/(2),(3)/(2),(1)/(2)`D. `(1)/(2),(5)/(2),(3)/(2)`

Given ` vecP=3 hati+2hatj+5hatk, veca=hati+hatj, vecb=hatj+hatk, vecc=

1.	Given ` vecP=3 hati+2hatj+5hatk, veca=hati+hatj, vecb=hatj+hatk, vecc=hati+hatk and vecP=xveca+yvecb+zvecc`, then x,y,z are respectivelyA. `(3)/(2), (1)/(2), (5)/(2)`B. `(1)/(2),(3)/(2),(5)/(2)`C. `(5)/(2),(3)/(2),(1)/(2)`D. `(1)/(2),(5)/(2),(3)/(2)`
Answer» Correct Answer - B `p=xa+yb+zc` `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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