Question

If A = 4i + 3j and B = 3i+ 4j, then cosine of the angle between A and (A + B) is what pls solve and give relevant answers... irrelevant answers would be reported.​

Answer 1

Question is majorly about the concept based on dot product of vectors. If you have two vectors A and B and you want to find the angle between them :

A • B = |A| |B| cosθ

A•B/ |A| |B| = cosθ

And if you want to find the angle between them,

θ = cos^(-1) A•B/ |A| |B|

(The bold letter represent vector here, because i don't know how to bring that arrow vector notation.)

So,using the above said things,we will solve this question. But first we need to find the magnitude of A and A + B. Magnitude of A :

A² = 4² + 3²

A² = 16 + 9

A² = 25

A = √25

$\sf\bold{A =5}$

∴ magnitude of A is 5.

Now,we need the magnitude of A + B.

Now you need to find A + B,

A + B = 4i + 3j + 3i + 4j

A+ B = 4i + 3i + 3j + 4j

A + B = (4+3)i + (3+4)j

$\tt\bold{A + B = 7i + 7j }$

Now let's find the magnitude of A + B :

A² + B² = 7² + 7²

A² + B² = 2 (7²)

A² + B² = 2 × 49

A² + B² = 98

$\sf\bold{A + B = √98}$

∴ magnitude of A + B is √98.

Now,we only need one thing and that is the dot product of vector A and A + B :

A • (A+B) = (4i + 3j) • (7i + 7j)

A• (A+B) = 4i (7i + 7j) • 3j (7i + 7j)

A• (A+B) = 28 + 0 • 0 + 21 [the dot product of i and j is 0 whereas dot product of same unit vectors is 1]

$\sf\bold{A•(A+B) = 28 + 21 = 49 }$

Now just put these values in the formula,

cosθ = A•(A+B)/|A| |A+B|

cosθ = 49/ 5 × √98

cosθ = 49/5 × 7√2

cosθ = 7/5 × √2

Now you can leave here or substitute the value of √2,

cosθ = 7/ 5 × 1.414

cosθ = 7/7.07

cosθ = 0.99

$\sf\bold{cosθ = 1}$