Question

If `6hati + 10hatj + 3hatk = x(hati + 3hatj + 5hatk) + y(hati - hatj + 5hatk) + z(hati + 3hatj - 4hatk)`, then ______

Options

• x = 1, y = 2, z = 3

• x = 3, y = 4, z = 1

• x = 1, y = 1, z = -3

• x = 2, y = -3, z = 5

Answer 1

If `6hati + 10hatj + 3hatk = x(hati + 3hatj + 5hatk) + y(hati - hatj + 5hatk) + z(hati + 3hatj - 4hatk)`, then x = 1, y = 2, z = 3.

Explanation:

`6hati + 10hatj + 3hatk = x(hati + 3hatj + 5hatk) + y(hati - hatj + 5hatk) + z(hati + 3hatj - 4hatk)`

⇒ `6hati + 10hatj + 3hatk = (x + y + z)i + (3x - y + 3z)hatj + (5x + 5y - 4z)hatk`

Comparing the co-efficients of `hati, hatj`and `hatk`, we get

x + y + z = 6

3x - y + 3z = 10

5x + 5y - 4z = 3

Solving the above equations , we get

x = 1, y = 2, z = 3