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

Question

51154 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-05T18:16:07

#v_1*v_2=-29#

I'm going to name these two vectors as #v_1# and #v_2#, where #v_1=2i-3j+4k=<2,-3,4># and #v_2=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)#.

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=-29#.

Answer 2 · 2018-03-05T18:16:48

The dot product is #=-29#

The dot product of #2# vectors

#veca= < x_1, y_1,z_1> #

and

#vecb= < x_2, y_2,z_2 >#

is

#veca.vecb = < x_1, y_1,z_1> . < x_2, y_2,z_2 >#

#=x_1x_2 +y_1y_2+z_1z_2#

Here, we have

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

#=2-3-28#

#=-29#