What is the dot product of (2i−3j+4k) and (i+j−7k)?

2 Answers
thelittlestleafeon
Mar 5, 2018

v1⋅v2=−29

Explanation:

I'm going to name these two vectors as v1 and v2, where v1=2i−3j+4k=<2,−3,4> and v2=i+j−7k=<1,1−7>.

The dot product of two vectors is defined as v1⋅v2=||v1||||v2||cos(θ)=(i1)(i2)+(j1)(j2)+(k1)(k2).

We don't have an angle to use, so we'll calculate the dot product using by adding the products of the components.

Therefore, v1⋅v2=(2)(1)+(−3)(1)+(4)(−7)=2−3−28=−29.

Narad T.
Mar 5, 2018

The dot product is =−29

Explanation:

The dot product of 2 vectors

→a=<x1,y1,z1>

and

→b=<x2,y2,z2>

is

→a.→b=<x1,y1,z1>.<x2,y2,z2>

=x1x2+y1y2+z1z2

Here, we have

<2,−3,4>.<1,1,−7=(2)⋅(1)+(−3)⋅(1)+(4)⋅(−7)

=2−3−28

=−29

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