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Q.

Maximum and minimum magnitude of resultant of vectors P→ and Q→ are 7 and 1 respectively. Then the maximum possible value of  P→ x Q→ is

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a

8

b

12

c

18

d

24

answer is B.

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Detailed Solution

P+Q = 7 and P-Q = 1. Solving P = 4, Q = 3∴P→ x Q→max=P.Q.sin90o=4 x 3 =12
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Maximum and minimum magnitude of resultant of vectors P→ and Q→ are 7 and 1 respectively. Then the maximum possible value of  P→ x Q→ is