Question

Answer this question Please​

Answer 1

Answer:

Step-by-step explanation:

$a=i+j+k\\\\b=i+2j+3k\\\\a+b=2i+3j+4k\\a-b=-j-2k\\\\let \;the\; vector\; perpendicular\;to\;a+b\;and\;a-b\;be c\\\\c=xi+yj+zk\\\\|\left[\begin{array}{ccc}x&y&z\\2&3&4\\0&-1&-2\end{array}\right]|=0\\ \\x(-2)-y(-4)+z(-2)=0\\\\x-2y+z=0\\\\now\\\\(a+b).c=0\\\\(2i+3j+4k)(xi+yj+zk)=0\\\\2x+3y+4z=0\\\\(a-b).c=0\\\\(-j-2k)(xi+yj+zk)=0\\\\y+2z=0\\\\2x+y+2y+4z=0\\\\2x+y+2(y+2z)=0\\\\2x+y=0$

We have

    2x+y=0

(+) y+2z=0

2x+2y+2z=0

x+y+z=0

On solving these equations

we have

x = 1

y = -2

z = 1

Therefore the vector is

i-2j+k

since it is a unit vector divide with its modulus

$\frac{i-2j+k}{3}$

Hope it helps

Answer 2

Answer:

Step-by-step explanation:a_×b_=i^(3-2) +j^(3-1) +k^(2-1) =a_×b=i^+3j^+k^=√6 and, unit vector perpendicular a and b is equal to a into B upon |A |into| B |is equal to i^ upon root 6 2i ^up on root 6 k^upon root 6