three vectors a,b and c are given as.?

Question

a = xi-3j+9k
b=-3i+xj-6k
c=9i-6j+xk

Find the largest value of x for which the magnitude of the resultant of the three vectors is equal to 17.

Answer 1

You want the maximum of the length of a+b+c:

|| a+b+c|| , expand this , differentiate, set to zero and solve for x.

For convenience, you can just use ||a+b+c||^2 rather than ||a+b+c||. this simplifys the calculation

Answer 2

Resultant
= a + b + c
= xi-3j+9k-3i+xj-6k+9i-6j+xk
= (x-3+9)i + (x-3-6)j + (x+9-6)k
= (x+6)i + (x-9)j + (x+3)k

Magnitude = 17
sqrt[(x+6)^2 + (x-9)^2 + (x+3)^2] = 17
x^2 + 12x + 36 + x^2 - 18x + 81 + x^2 + 6x + 9 = 17^2
3x^2 + 126 = 289
3x^2 = 163
x^2 = 54 1/3
x = 7.37, -7.37
Hence the largest value of x is 7.37.