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

2 Answers
thelittlestleafeon
Mar 5, 2018

v_1*v_2=-29v1⋅v2=−29

Explanation:

I'm going to name these two vectors as v_1v1 and v_2v2, where v_1=2i-3j+4k=<2,-3,4>v1=2i−3j+4k=<2,−3,4> and v_2=i+j-7k=<1,1-7>v2=i+j−7k=<1,1−7>.

The dot product of two vectors is defined as v_1*v_2=||v_1|| ||v_2|| cos(theta)=(i_1)(i_2)+(j_1)(j_2)+(k_1)(k_2)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, v_1*v_2=(2)(1)+(-3)(1)+(4)(-7)=2-3-28=-29v1⋅v2=(2)(1)+(−3)(1)+(4)(−7)=2−3−28=−29.

Narad T.
Mar 5, 2018

The dot product is =-29=−29

Explanation:

The dot product of 22 vectors

veca= < x_1, y_1,z_1> →a=<x1,y1,z1>

and

vecb= < x_2, y_2,z_2 >→b=<x2,y2,z2>

is

veca.vecb = < x_1, y_1,z_1> . < x_2, y_2,z_2 >→a.→b=<x1,y1,z1>.<x2,y2,z2>

=x_1x_2 +y_1y_2+z_1z_2=x1x2+y1y2+z1z2

Here, we have

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

=2-3-28=2−3−28

=-29=−29

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