Question

find the angle between the vectors 5i+3j+4k and 6i-8j-k.

Answer 1

It may be this......
A=5i+3j+4k
B=6i-8j-k
A=root 5square+3square+4square
= root 50
B=root 6square +(-8square)+(-1square)
=root 101
A.B=(5i+3j+4k).(6i-8j-k)
=2
angle between A and B= cos theeta
=A.B/AB
2/root50 root101

Answer 2

Answer:

the angle between θ =  $cos^{-1} ( \frac{\sqrt{2} }{5\sqrt{101} } )$

Step-by-step explanation:

Given Problem

find the angle between the vectors 5i+3j+4k and 6i-8j-k

let the given vectors are  vector a = 5i+3j+4k

                                          vector b = 6i-8j-k

the angle between 2 vectors  

θ = cos⁻¹ [dot product of 2 vector ÷ magnitude of 2 vectors ]  

now find product and magnitude of two vectors

product of the vector

= ( 5i+3j+4k )  ( 6i-8j-k )  

= ( 5×6 + 3×(-8) + 4×(-1) )  

=   30 - 24 - 4 = 2

⇒ product of given vectors = 2

Magnitude of vectors  

magnitude of 5i+3j+4k  = $\sqrt{5^{2} + 3^{2} + 4^{2} }$  

                                       =  $\sqrt{25 + 9 + 16 }$

                                       = $\sqrt{50}$  =  $\sqrt{25(2) }$ = 5 √2  

magnitude of  6i-8j-k  =  $\sqrt{6^{2}+ (-8)^{2} + (-1)^{2} }$  

                                    =  $\sqrt{36 + 64 + 1 }$  

                                    =  $\sqrt{101}$  

angle between 2 vectors  θ = $cos^{-1} ( \frac{a . b}{|a||b|})$  

                                              =  $cos^{-1} ( \frac{2}{5 \sqrt{2} (\sqrt{101} } )$

                                              =  $cos^{-1} (\frac{\sqrt{2} \sqrt{2} }{5\sqrt{2} \sqrt{101} } )$

                                           θ  = $cos^{-1} (\frac{\sqrt{2}}{5\sqrt{101} } )$