Question

.5: If a = i - j , b = i+j+k and c be a vector such that a x c + b = o and a.c= 4, then IcI^2
is equal to_

Answer 1

Answer:

Products 1 - 8 · This brings us to the calculus of several variables. .... A+B= PR=C+D. Then V =+A+$Bequals W=+C++D. ... The combinations of i and j (or i, j, k) produce all vectors v.

Answer 2

IcI^2  is equal to 9.5.

• Let c=xi+yj+zk. Now it is given that a×c + b = 0 and a.c=4

• Now calculating a×c we get a×c = $\left[\begin{array}{ccc}i&j&k\\1&-1&0\\x&y&z\end{array}\right]$ = -zi-zj+(x+y)k

• Adding it with b vector we get  a×c + b = (1-z)i+(1-z)j+(x+y+1)k = 0

• Hence we can assume individual components are 0 , Hence from that we get z = 1

• From there we also get x+y = -1
• Now calculating a.c, we get a.c = x-y = 4
• Now after solving these two equations we get x = 1.5 and y = -2.5
• Hence c= 1.5i-2.5j+k
• Hence IcI^2 = (1.5)^2+(-2.5)^2+1^2 = 9.5