Question

If a=i+j+k and b=j-k find a vector c such that axc=b and



a.C=3

Answer 1

Answer:

c= 1/3[4i+4j+2k]

Step-by-step explanation:

a=i+j+k

b=j-k

also axc-b= 0 ( manupulation )

now post multiply (cross) by a

ax(axc) -axb=0

also axb= -i-2j-k ( determinant form )

so (a.c)a-(a.a)c - axb = 0 (using identity)

3a-3c +i+2j+k =0

3a +i+2j+k=3c

3i+3j+k+i+2j+k =3c

4i+4j+2k=3c

c= 1/3[4i+4j+2k]

hope it helped.....

Answer 2

The vector c = 5/3i+2/3j+2/3k

• Let c = xi+yj+zk. Now it is given that a×c = b and a.c=3
• Now calculating a×c we get a×c = $\left[\begin{array}{ccc}i&j&k\\1&1&1\\x&y&z\end{array}\right]$ = (z-y)i+(x-z)j+(y-x)k = j - k
• Hence we can assume individual components are equal , Hence from that we get z-y = 0, x-z = 1, x-y = 1
• This implies z = y and x=y+1
• Now calculating a.c, we get a.c = x+y+z = 3
• Now after putting previous values we get y+1+y+y = 3. Hence y = 2/3
• Hence x = 5/3 and z = 2/3
• Hence the required vector c = 5/3i+2/3j+2/3k