Question

find the angle between 2i-j+k and i-j+0k

Answer 1

Answer:

cos inverse=(-1/3) is the angle

Answer 2

Given :-

◉ Vectors,

• A = 2i - j + k
• B = i - j + 0k

To Find :-

◉ Angle between the vectors A and B.

Solution :-

First let us find the resultant vector by applying the cross product on A and B,

| i j k |

⇒ A × B = | 2 -1 1 |

| 1 -1 0 |

Opening the determinant,

⇒ A × B = (-1×0 - 1×-1)i - (2×0 - 1×1)j + (2×-1 - 2×-1)k

⇒ A × B = i + j + 0k

⇒ |A × B| = √(1² + 1² + 0²)

⇒ |A × B| = √2

Now, Let us find the magnitude of A and B,

⇒ |A| = √(2² + (-1)² + 1²)

⇒ |A| = √(4 + 2)

⇒ |A| = √6

Similarly, Magnitude of vector B,

⇒ |B| = √(1² + (-1)²)

⇒ |B| = √(1 + 1)

⇒ |B| = √2

We know,

⇒ A × B = |A| |B| sin θ

⇒ sin θ = |A × B| / |A| |B|

⇒ sin θ = √2 / √6 × √2

⇒ sin θ = 1/√6

⇒ θ = sin⁻¹ 1/√6

Hence, The angle between the two given vectors is sin⁻¹ 1/√6

Some Information :-

☛ Cross product of two vectors gives a vector.

While, Scalar product or dot product gives a scalar , generally a number.

• Scalar Product = |A| |B| cos θ
• Cross Product = |A| |B| sin θ