Let u=5i−j, v=3i+j, w=i+3j Find the specified scalar. (4u)•v

Answer 1

The fist thing to understand is that i*i = 1, j*j =1, i*j = 0, j*i =0. Once you know that you can do the problem like you would do any other multiplication one.

(4u)*v

= (4*(5i -j))*(3i+j)

= (20i - 4j) *( 3i + j )

= 60i*i + 20 i*j - 12 j*i - 4j*j

Only the i*i and j*j terms are non-zero therefore

60-4 = 56

Now a bit more on why i*i = 1, j*j =1, i*j = 0, j*i =0

The scalar multiplication is more general than this, but from the perspective of physics you can think about it like so. The operator (a*b) is the one that gives you the projection of vector "a" with respect to the vector "b".

The vectors i and j are special vectors that have length 1 and are perpendicular to one another. So when we do this i*i we are asking what is the projection of vector i with it self. Which is just the length of i therefore 1. Similarly with j, now let's look at i*j. Since i and j are perpendicular the projection of each other with respect to the other is zero.

Answer 2

Question: Let u = 5i - j, v = 3i + j, and w = i + 3j. Find the specified scalar: (4u)•v.

Answer: The dominant operation here is a dot product, so the final answer is a scalar. Let's find 4u first, and then dot that vector with v.

I prefer to work in component form. Let me make the cosmetic update here:

u = <5, -1, 0>

v = <3, 1, 0>

w = <1, 3, 0>

Let's do 4u:

4u = 4<5, -1, 0> = <4(5), 4(-1), 4(0)> = <20, -4, 0>

Let's take that vector and dot it with v:

<20, -4, 0>•<3, 1, 0>

=20(3) + (-4)(1) + 0(0)

=60 - 4 + 0

=56

(4u)•v = 56.