Question

Please solve this question ​

Answer 1

Answer:

$\huge{\mathcal{\purple{A}\green{n}\pink{s}\blue{w}\purple{E}\green{r}}}$

If the length of a⃗ , b⃗ and a⃗ +b⃗ are the same, we must have:

|a⃗ +b⃗ |2=|a⃗ |2

This means:

(a⃗ +b⃗ )⋅(a⃗ +b⃗ )=|a⃗ |2⇔|a⃗ |2+|b⃗ |2+2a⃗ ⋅b⃗ = |a⃗ |2

Since |a⃗ |2 and |b⃗ |2 are equal, this reduces to:

|a⃗ |2+2a⃗ ⋅b⃗ =0

By the law of cosines a⃗ ⋅b⃗ =|a⃗ ||b⃗ |cosθ , where θ is the angle between the vectors. Here, this reduces to a⃗ ⋅b⃗ =|a⃗ |2cosθ , and we get:

|a⃗ |2+2|a⃗ |2cosθ=0⇔|a⃗ |2(1+2cosθ)=0

Isolating, we get:

cosθ=−12

This has the solution of θ being equal to 120 degrees.

Answer 2

Answer:

➡If the length of a⃗ , b⃗ and a⃗ +b⃗ are the same, we must have:

|a⃗ +b⃗ |2=|a⃗ |2

➡This means:

(a⃗ +b⃗ )⋅(a⃗ +b⃗ )=|a⃗ |2⇔|a⃗ |2+|b⃗ |2+2a⃗ ⋅b⃗ = |a⃗ |2

➡Since |a⃗ |2 and |b⃗ |2 are equal, this reduces to:

|a⃗ |2+2a⃗ ⋅b⃗ =0

By the law of cosines a⃗ ⋅b⃗ =|a⃗ ||b⃗ |cosθ , where θ is the angle between the vectors. Here, this reduces to a⃗ ⋅b⃗ =|a⃗ |2cosθ , and we get:

|a⃗ |2+2|a⃗ |2cosθ=0⇔|a⃗ |2(1+2cosθ)=0

Isolating, we get:

cosθ=−12

This has the solution of θ being equal to 120 degrees.

hope help u mate ✌