IIT JEE , AIEEE Forum - Mathematics , Vectors

karthik_abiram

Q] A force F= i -8j-7k is resolved along three mutually

perpendicular directions ,one of which is in the direction

of vector "a" =2i+2j+k.Then the component of F vector

in the direction of "a" vector is

1] -14i -14j-7k

2]-7/3 (2i+2j+k)

3] 1/3(-2i-2j-k)

4] 7/3 (2i+2j+k)

kghedriu

Consider two vectors a and b...

projection of a on b = a.b/[b] where [x] means modulus of x... and a.b is the dot product...

the projected vector is therefore given by a.b^ where b^ is the unit vector along the direction of b........

in the present situation, we need F along a...

so, the reqd vector is :F.a^

and we know that a^ = a/[a]

hence find out the answer now...

we have [F]=

114

and angle between the vectors = F.a/[F][a] = -7/

114...

so, the projection is -7/3(2i+2j+2k)

option (2)

hope m clear..

thank u..

Goutham Harsha.K

rohitdharmik

projection of vector f along a is given by (f.a/[a])a
therfore,(i-8j-7k.2i+2j+k/[2i+2j+k])2i+2j+k
=(2-16+7/3)2i+2j+k
=-7/32i+2j+k

karthik_abiram

thank you all