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Two vectors have magnitudes 6 and 8 unit...

Two vectors have magnitudes `6` and `8` units, respectively. Find the magnitude of the resultant vector if the angle between vectors is (a) `60^(@)` (b) `90^(@)` and ( c ) `120^(@)`.

Text Solution

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To solve the problem of finding the magnitude of the resultant vector for the given angles between two vectors with magnitudes 6 and 8 units, we will use the formula for the magnitude of the resultant vector: \[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \] where: - \( R \) is the magnitude of the resultant vector, ...
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Two vectors A and B have magnitudes 6 units and 8 units respectively. Find |A-B|, if the angle between two vectors is .(a) 0^(@) (b) 180^(@) (c) 180^(@) (d) 120^(@) .

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Knowledge Check

  • When two vectors vecA and vecB of magnitudes a and b respectively are added, the magnitude of resultant vector is always

    A
    Equal to `(a+b)`
    B
    Less than `(a+b)`
    C
    Greater than `(a+b)`
    D
    Not greater than `(a+b)`
  • The magnitude of two vectors are 16 and 12 units respectively and the magnitude of their scalar product is 98sqrt2 units. The angle between the vectors would be

    A
    30°
    B
    45°
    C
    60°
    D
    90°
  • If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is

    A
    `45^(@)`
    B
    `180^(@)`
    C
    `0^(@)`
    D
    `90^(@)`
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