Question

The force F = (i +bj+3k)N is rotated through an angle a , then it becomes {2i (2b-1)j+k}N. The value of b is

Answer 1

Answer:

Explanation: According to the question,by rotating vector the vector will remain same.

Therefore we will equate the magnitude of the two vectors.

√1²+b²+3²=√2²+(2b-1)I+1²

Or,1+b²+9=4+(2b-1)I+1[squaring both sides]

Or,b²+10=5+4b²+1-4b

Or,3b²-4b-4=0

Factorizing by discriminant method:

b=-(-4)+√(-4)²-4*3*-4/2*3

b=4+√16+48/6=4+8/6

Or b=2

For negative value of b:

b=4-√16+48/6

=-2/3

Therefore,b has two values.b=

2 or -2/3

Answer 2

Answer:

The value of b is $2$ or $-2/3$.

Explanation:

Given the force $F_1 = (\hat i +b\hat j+3\hat k)N$

Rotated through an angle $a$

Final force, $F_2 = (2\hat i +(2b-1)b\hat j+\hat k)N$

Rotating a vector about an angle changes its direction keeping its magnitude constant.

Therefore, both the forces remain the same.

$|\vec F_1|=|\vec F_2|$

$\implies |\hat i +b\hat j+3\hat k |=| 2\hat i +(2b-1)b\hat j+\hat k|$

$\implies \sqrt{1^2+b^2+3^2} =\sqrt{2^2+(2b-1)^2+1^2}$

squaring on both sides

$1^2+b^2+3^2 =2^2+(2b-1)^2+1^2$

$\implies 1+b^2+9 =4+4b^2-4b+1+1$

$\implies b^2+10=4b^2-4b+6$

$\implies 3b^2-4b-4=0$

For any quadratic equation $ax^2+bx+c=0$, the quadratic formula gives the root is given by

$x=\dfrac{-b\pm\sqrt{b^{2} -4ac} }{2a}$

Using the values from equation (1),

$b=\dfrac{-(-4)\pm\sqrt{(-4)^{2} -4\times 3\times (-4)} }{2\times 3}$

  $=\dfrac{4\pm\sqrt{16 +48} }{6}=\dfrac{4\pm\sqrt{64} }{6}$

  $=\dfrac{4\pm8 }{6}$

Thus, the roots are given by

$b=\dfrac{4+8 }{6}~\mbox{or}~\dfrac{4-8 }{6}$

  $=\dfrac{12 }{6}~\mbox{or}~\dfrac{-4 }{6}$

  $=2~\mbox{or}~\dfrac{-2 }{3}$

Therefore, the value of b is $2$ or $-2/3$.