IIT JEE , AIEEE Forum - Mathematics , Vectors

luxmi

if a+b+c=0 and |a|=3 |b|=5 |c|=5, then the angle btwn a and b is?

a) 45

b) 30

c) 60

d) 90

aditya_arora04

a+b+c=0

a+b = -c

squaring both sides

|a|² + |b|² + 2|a||b|cost = |c|²

put the values to get t

luxmi

but we r gettin cos

= -3/10

raulrag009

since a+b+c=0;
therefore it will be a triangle

then use cosine formulae to solve

luxmi

still we get da same

cost=5^2 +3^2 -5^2/30

=3/10

karthik_abiram

r u sure that both |b|=|c|=5 ?

luxmi

yes they both r de same

|b|=|c|=5

karthik_abiram

Then the ans is cos @ = - 3/10

There must propably have been a typing mistake in the book.........

raulrag009

yes the ans is -3/10

karthik_abiram

If | b | = 4 ,then @ = 90

Similar Qn in my school book........but with diff choices........and in that |b | = 4 and answer is 90

cool_gal

i think ur ques is wrong

check it again .

with ur data the ans is coming out to be -3/10.

gokulkrishna

a + b = - c

a²+ b^{2 _{+ 2a.b = c}2}

^{9 + 25 +30cos = 25}

^{cos = - 9/30 = - 3/10}

feynmann

ans is pi - arccos ( 3/10 )

Got it , Or need explanation .?

approx 107 . 456 deg

drish

yeah kartik is right this same ques (where|b|=4)is given in tata mcgrath hill .in addition to kartik's reply i would like to say tat output of cos can never be (-)ve.

luxmi

why shouldnt it be negative? i mean wen the angle is obtuse or sumthin

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