Question

Given the vectors
A = 2i + 3j-k: B = 3-2-2k & C = pi + pj+2pk find the angle between (A-B)&C​

Answer 1

Answer: let a -b vector be p vector

Explanation:

Answer 2

Answer:

The Angle between (A-B) and C is 40°

Explanation:

Given : A = 2i + 3j - k     B= 3i - 2j - 3k      C= pi + pj + 2pk

Step 1 : to  finding vector A - B  

A - B = (2i +3j -k) -(3i - 2j - 3k )  

 A - B =  (-i + j - 4k)

Step 2:  to find dot product of (A-B) and C

 (A-B).C = (-i +j -4k) .(pi + pj + 2pk)      

       = -p + p- 8p  (A-B) .C

       = -8p

Step 3 : now we find magnitude of A-B and C

A-B = $\sqrt{(-1)^{2}+1^{2} +(-4)^{2} }$        

 = $\sqrt{18}$

C= $\sqrt{p^{2}+p^{2}+(2p)^{2} }$      

=$\sqrt{6p^{2} }$

Step 4: now we can also write

 (A-B).C =(A-B)(C) cos θ

 $-8p=\sqrt{18} \sqrt{6p^{2} }cos \theta$

 $-\frac{8}{6\sqrt{3} } = cos\theta$

 $cos \theta = 0.77$

 $\theta = 40$°