In this page, you will learn about the vector dot product with the scalar multiple of a vector.

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Given `vec p = 2i+j-k` and `lambda = 3.1` what is the scalar multiple `3.1 vec p`?

- `6.2i+3.1j-3.1k`
- `6.2i+3.1j-3.1k`
- `6.2i+j-k`

The answer is '`6.2i+3.1j-3.1k`'

Given the definition of dot product as

`vec p cdot vec q = |vec p||vec q|cos theta`

What is `vec p cdot (lambda vec q)`?

- `lambda(vec p cdot vec q)`
- `|p||lambda q|cos(theta)` if `lambda` is positive
- `lambda|p||q|cos(theta)`
- all the above
- all the above

The answer is 'All the above'

Note that if `lambda` is a negative number, it equals `|p||lambda q|cos(180+theta)`.

Given the definition of dot product as

`vec p cdot vec q = p_xq_x+p_yq_y+p_zq_z`

What is `vec p cdot (lambda vec q)`?

- `lambda (vec p cdot vec q)`
- `p_x(lambda q_x)+p_y(lambda q_y)+p_z(lambda q_z)`
- `lambda (p_xq_x+p_yq_y+p_zq_z)`
- all the above
- all the above

The answer is 'All the above'

• Dot product with a scalar multiple of a vector equals the scalar multiple of the dot product with the vector.

**Dot Product of scalar multiple: ** For any vectors `vec p, vec q`

`vec p cdot (lambda vec q) = lambda (vec p cdot vec q)`

*Solved Exercise Problem: *

Given that `vec p cdot vec q = 3j+1.1k`, what is `vec p/(2.1) cdot vec q`?

- `3/(2.1)j+1.1/(2.1)k`
- `3/(2.1)j+1.1/(2.1)k`
- `6.3j+2.31k`
- `3/(2.1)j+1.1k`

The answer is '`3/(2.1)j+1.1/(2.1)k`'.

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