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Find magnitude and direction cosines of the vector, `A= (3hati - 4hatj+5hatk).`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`A or |A| = sqrt((3)^2 + (-4)^2 + (5)^2) = 5sqrt2 units`
Directions of cosines are,
`cos alpha = (A_x)/(A) = (3)/(5sqrt2`
`cos beta = (A_y)/(A) = (-4)/(5sqrt2) and `
`cos_gamma = (A_z)/(A) =(1)/(sqrt2)`
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Knowledge Check

  • The direction cosines of the vector 3hati-4hatj+5hatk are

    A
    `(3)/(5),(-4)/(5),(1)/(5)`
    B
    `(3)/(5sqrt(2)),(-4)/(5sqrt(2)),(1)/(sqrt(2))`
    C
    `(3)/(sqrt(2)),(-4)/(sqrt(2)),(1)/(sqrt(2))`
    D
    `(3)/(5sqrt(2)),(4)/(5sqrt(2)),(1)/(sqrt(2))`.

  • The direction cosines of the vector 3hati-4hatj+5hatk are

    A
    `(3)/(5),-(4)/(5),(1)/(5)`
    B
    `(3)/(5sqrt(2)),(-4)/(5sqrt(2)),(1)/(sqrt(2))`
    C
    `(3)/(sqrt(2)),(-4)/(sqrt(2)),(1)/(sqrt(2))`
    D
    `(3)/(5sqrt(2)),(4)/(5sqrt(2)),(1)/(sqrt(2))`.

  • The direction cosines of the vector 3hati+4hatk are

    A
    `(3)/(4),0,(4)/(5)`
    B
    `(3)/(5),(4)/(5),0`
    C
    `0,(3)/(5),(4)/(5)`
    D
    `(3)/(5),0,(4)/(5)`

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