Question

IfA=(2,-1, 5), B=(2, 1, 4) and C(1, 1, 0) then find a unit vector parallel to AC-BC​

Answer 1

Answer:

Correct option is

B

31(2i−j+2k)

a+b−c=i(2+1−5)+j(1−1+1)+k(−1+0−1)

                 =−2i+j−2k

 Unit vector along the above direction is

    =∣a+b−c∣a+b−c

    =4+1+4−2i+j−2k

    =3−2i+j−2k

Vector anti parallel to the above is

    =31(2i−j+2k)

Step-by-step explanation:

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Answer 2

Answer:

-j

Step-by-step explanation:

A=(2,-1, 5), B=(2, 1, 4) and C(1, 1, 0)

AC= ( 2.1 ,-1.1 ,5.0 ) = (2 , -1 , 0 )

BC= (2.1 ,1.1, 4.0 ) = ( 2, 1, 0)

AC - BC = ( 2-2, -1 -1, 0-0)

=(0, -2, 0)

=(0i , -2j , 0k)

Now we want unit vector parallel to AC-BC

so = (0i -2j +0k)/ root( 0^2 + (-2)^2 + 0^2 )

= -2j / root (4)

=-2j / 2

= -j

Keep Solving :-)