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Grade 11Mechanics

Given a vector A = 2i + 3j and a vector B = i + j.
The component of the vector A along B is
(a) 1/(2)½ (i + j)
(b) 3/(2)½ (i + j)
(c) 5/(2)½ (i + j)
(d) 7/(2)½ (i + j)

Rajesh

9 Years agoGrade 11

7 Answers

Vikas TU

9 Years ago

The component of the vector A along B geometrically is:

Bcos(theta) = (A.B/|B|).(Bcap)

=5/(2)½ (i + j)

option is correct.

Pushkar Ranjan

8 Years ago

A along B is given as” A cosθ = A.B/|B|”

which formulates to give 2+3 / √2 = 5/ √2

the option (c.) is correct

meet kumar

7 Years ago

As we know

ABcosø=A.B

From this eq we get

ACosø=A.B/|B|

by putting values from given data

A Cosø=5/root2

Soo C is the right option

M

7 Years ago

2square+3square whole root=5

1square+1square whole root=root 2

then multiply

both you could get needed answer

that is 5root2

Aahira

6 Years ago

The component of A along vector B=) A. B) where B^=B/b where b is the magnitude of vector B ...

Now do A. B

Thn find B^ wch is rdq soln

Rishi Sharma

6 Years ago

Hello students,
The problemis of vector and asked the component of A along B.
We know that,
the component of the vector A along B geometrically is :=(A.B).(Bcap)
=(A.B).(B/|B|)
=((2i +3j) . (i + j)) . ((i+j)/(2)½)
=(2+3) . ((i+j)/(2)½)
=5/(2)½ (i + j)
I hope this solution would solve your all doubts.
Thank You

LISA

5 Years ago

Magnitude of the component (let us say vecC )vector of A⃗ along B⃗ is given by,

|C⃗ |=A⃗ ∙B⃗ |B⃗ |=(2i^+3j^)∙(i^+j^)12+12√=2×1+3×12√=52√2

Since,

i^∙i^=1

j^∙j^=1

i^∙j^=0=j^∙i^

Again, the vector along B⃗ is given by

C⃗ =|C⃗ |× unit vector along B⃗

=|C⃗ |×B⃗ |B⃗ |

=52√2(i^+j^)2√

=52i^+52j^

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