Question

the angle between two vectors given by 6i+6j-3k and 7i+4j+4k is....

Answer 1

Answer:

Here

      a= 6i^+6j^-3k^  

       b=7i^+4j^+4k^

Formula -

    a.b =|a| |b| cos(x), here we need to find value of ‘x’

Explanation:

Firstly put given values in left side of formula a.b  

       a.b= 6 *7(i*i)+6*4(j*j)-3*4(k*k)

  L.H.S= 42+24–12

  L.H.S = 54

Then put given values in right side of formula |a| |b|  

         |a|= sq.rt (6*6+6*6+3*3), |b|= sq.rt (7*7+4*4+4*4)

         |a|= sq.rt (36+36+9), |b|= sq.rt (49+16+16)

         |a|= sq.rt (81), |b|= sq.rt (81)

R.H.S |a| |b|=9 *9= 81

So, put values of L.H.S and R.H.S into formula to get value of ‘x’

      a.b =|a| |b| cos(x)

   cos(x)=a.b/|a| |b|

    cos(x)= 54/81

    cos(x)=2/3

            x= cos^-1(2/3)

Answer 2

here is ur ans i hope it will help u