Question

1. The force F =(+bj+3k)N is rotated through an
angle a, then it becomes {2+(26 - 1)j + k}N
The value of bis :-

(1) 2
(2)-2/3
(3)2/3
(4) Both (1) and (2)​

Answer 1

Answer:

Both (1) and (2)

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

Your question:

1. 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 :-

1. 2
2. -2/3
3. 2/3
4. Both (1) and (2)

Solution :

Firstly, the magnitude of the vector will remain same even when we rotate the vector on its axis in a same plane. By rotating the vector, we're only changing its direction.

A/q

We will equate the prior and latter magnitude.

|i+bj+3k|=>|2i+(2b-1)j+k|

$\sf 1^2+b^2+3^2=2^2+4b^2+1^2-4b+1 \\\\\sf 10+b^2=4b^2-4b+6 \\\\\sf 4b^2-4b+6-10-b^2 \\\\\sf 3b^2-4b-4 =0 \\\\\sf 3b^2-6b+2b-4=0 \\\\\sf 3b(b-2)+2(b-2)=0 \\\\\sf (3b+2)(b-2) \\\\\sf So, b= \frac{-2}{3}~or~2$