Question

find the angle between vector a equals to 11 iCAP + 4 j cap + 2 k cap on x axis

Answer 1

Answer ⇒ The angle made by the vector A with the x-axis is Cos⁻¹(11/√141).

Explanation ⇒

       Given conditions, A = 11 i + 4j + 2k

Unit vector along 'A' = Vector A/Magnitude of vector A.

∴ Magnitude of vector A = √[(11)² +(4)² + (2)²]

= √[121 + 20]

= √141

Now,

Unit Vector Along A = [11 i + 4j + 2k]/√141

= (11/√141) i + (4/√141) j + (2/√141) k

Thus, angle which the vector A makes with x - axis can be given in terms of its Cosines as,

Cosθ = 11/√141

∴ θ = Cos⁻¹(11/√141)

Thus, the angle made by the vector A with the x-axis is Cos⁻¹(11/√141).

Hope it helps.

Answer 2

Explanation ⇒

       Given conditions, A = 11 i + 4j + 2k

Unit vector along 'A' = Vector A/Magnitude of vector A.

∴ Magnitude of vector A = √[(11)² +(4)² + (2)²]

= √[121 + 20]

= √141

Now,

Unit Vector Along A = [11 i + 4j + 2k]/√141

= (11/√141) i + (4/√141) j + (2/√141) k

Thus, angle which the vector A makes with x - axis can be given in terms of its Cosines as,

Cosθ = 11/√141

∴ θ = Cos⁻¹(11/√141)

Thus, the angle made by the vector A with the x-axis is Cos⁻¹(11/√141).