Question

cosines of angle made by a vector with X,Y axes are 3/5√2,4/5√2 respectively.if the magnitude of the vector is 10√2,then the vector is
A) 8i+6j-10k
B) 6i-8j-10k
C) -6i-8j+10k
D) 6i+8j+10k​

Answer 1

Answer:

8i+6j-10k is answering of the

Answer 2

Given info : cosines of angle made by a vector with X,Y axes are 3/5√2,4/5√2 respectively.

To find : if the magnitude of the vector is 10√2 , then the vector is ..

solution : we know, cosine angles are given by,

cos²α + cos²β + cos²γ = 1

here, cosα = 3/5√2 and cosβ = 4/5√2

so (3/5√2)² + (4/5√2)² + cos²γ = 1

⇒9/50 + 16/50 + cos²γ = 1

⇒cos²γ = 1 - 25/50 = 1 - 1/2 = 1/2

⇒cosγ = ± 1/√2 = ± 5/5√2

we also know,

cosα = x/√(x² + y² + z²) , cosβ = y/√(x² + y² + z²) and cosγ = z/√(x² + y² + z²)

so a unit vector is , r = cosα i + cosβ j + cos γ k

now, the vector is written by, A = r|A|

= (cosα i + cosβ j + cos γ k) 10√2

= (3/5√2 i + 4/5√2 j ± 5/5√2 k) × 10√2

= 6i + 8j ± 10k

Therefore the vector is either 6i + 8j - 10k or 6i + 8j + 10k.