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Question

Find a unit vector perpendicular to each of the vectors (a+b) and (ab), where a=^i+^j+^k,
b=^i+2^j+3^k

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Solution

Given:a=i+j+k and b=i+2j+ 3k

(a+b)= (^i+^j+^k)+(^i+2^j+3^k)

(a+b)=2^i+3^j+4^k

(ab)=(^i+^j+^k)(^i+2^j+3^k)

ab=0^i^j2^k

Let c=(a+b)×(ab)

c=∣ ∣ ∣^i^j^k234012∣ ∣ ∣

c=^i[(3×2)(1×4)]^J[(2×2)(0×4)]+^k[(2×1)(0×3)]

c=^i[6(4)]^j[40]+^k[20]

c=2^i+4^J 2^k

Unit vector c=cmagnitude of c

^c=(2^i+4^j2^k)(2)2+(4)2+(2)2

^c=(2^i+4^j2^k)4+16+4

^c=(c2^i+4^j2^k)26

Unit vectorc(^c)=16^i+26^j16^k

Hence, the unit vectorcis c=16^i+26^j16^k










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