Question

Find a vector of magnitude 3 in the direction of v=4i-3k

Answer 1

To find a vector of magnitude 3 in the direction of v = 4i - 3k, first calculate the unit vector by dividing v by its magnitude, and then multiply the unit vector by 3 to get the desired vector with magnitude 3.

Step-by-step explanation:

To find a vector of magnitude 3 in the direction of v = 4i - 3k, we must first find the unit vector in the direction of v. A unit vector has a magnitude of 1 and points in the same direction as the original vector. To find this unit vector, divide each component of v by its magnitude.

The magnitude of v is obtained using the formula
`\(|v| = \sqrt{v_(x)^(2) + v_(y)^(2) + v_(z)^(2)}\)`, where vx, vy, and vz are the components of the vector v. For the vector v = 4i - 3k, the magnitude is
`\(|v| = \sqrt{4^(2) + 0^(2) -3^(2)} = √(16 + 9) = √(25) = 5\).`

Next, the unit vector u in the direction of v is u = (1/|v|) * v = (1/5)(4i - 3k) = 0.8i - 0.6k.

To get the desired vector of magnitude 3, we multiply the unit vector by 3: 3u = 3(0.8i - 0.6k) = 2.4i - 1.8k. Thus, the vector 2.4i - 1.8k has a magnitude of 3 and is in the direction of the original vector v.

Answer 2

`(12i)/(5) - (9k)/(5)`

Step-by-step explanation:

The magnitude of a vector v = ai + bk is

`|v| = \sqrt{a^(2) + b^(2)`

In this problem, we have that:

Find a vector of magnitude 3 in the direction of v=4i-3k:

The first step is finding the unit vector of v,
`v_(u)`.

So

`|v| = \sqrt{4^(2) + (-3)^(2)} = 5`

`v_(u) = (4i)/(5) - (3k)/(5)`

Magnitude 3, same direction.

We multiply the unit vector by +3, since it is in the same direction. If it was in the opposite direction, we would have multiplied by -3.

The answer is:

`(12i)/(5) - (9k)/(5)`