Prove that triangle awm is congruent to triangle cpm using a two column proof

ΔAWM ≅ ΔCPM proved

Step-by-step explanation:

See the given diagram with this question.

We have to prove ΔAWM ≅ ΔCPM

Now, (i) M is the mid point of AC, so AM =MC

(ii) Now, W is the mid point of AB, hence, AW = WB =

Again, P is the mid point of BC i.e

Now, it is given that BP = PM = MW = WB

Hence, BP = WB , hence, AW = PC

(iii) Again, PM = MW

Therefore, from the SSS rule, it is proved that ΔAWM ≅ ΔCPM. (Answer)

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

a

step-by-step explanation: