# Suppose vector 1, vector 2, vector 3 is a set of vectors mutually perpendicular. Assume that Length of vector v1 = sqrt(35) length of vector 2 = sqrt(94) length vector v3 = sqrt(36) Let w be a...

Suppose vector 1, vector 2, vector 3 is a set of vectors mutually perpendicular. Assume that

Length of vector v1 = sqrt(35)

length of vector 2 = sqrt(94)

length vector v3 = sqrt(36)

Let w be a vector in Span(v1,v2,v3) such that

(vector w)dot product(vector v1) = 35

(vector w)dot product(vector v2) = -188

(vector w)dot product(vector v3) = 36

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As W is a vector in span (V_1, V_2, V_3, V_4).

Also given mod(V_1)=sqrt(35)

mod(V_2)=sqrt(94)

mod(V_3)=sqrt(36)

Now w can be written as w=projection of w on V_1+projection of w on V_2+projection of w on V_3

So, projection of w on V_1={(w.V_1)/(mod(V_1))^2}.V_1

or, projection of w on V_1=(35/35)V_1=1V_1

Simlarly projection of w on V_2=(-188/94)V_2=-2V_2

and projection of w on V_3=(36/36)V_3=1V_3

so, w=V_1-2V_2+V_3

Answer.

Let vectors are

`v_1=(a,0,0)`

`v_2=(0,b,0)`

`v_3=(0,0,3)`

so lenth of vector `v_1=sqrt(35)`

length of vector `v_2=sqrt(96)` and

length of vector `v_3=sqrt(36)`

Thus we have a=sqrt(35) ,b=sqrt(96) and c=sqrt(36)

Let vector `w=(x,y,z)`

By given conditions, we have

`w*v_1=x*a`

`=x*sqrt(35)=35`

`x=sqrt(35)` (i)

`w*v_2=y*b`

`y*sqrt(94)=-188`

`y=-2sqrt(94)` (ii) and

`w*v_3=z*c`

`z*sqrt(36)=36`

`z=sqrt(36)`

So vectot `w=(sqrt(35),-2sqrt(94),sqrt(36))`