Question

The vector i + xj + 3k is rotated through an angle theta and doubled in magnitude, then it becomes

4i + (4x – 2) j + 2k. The values of x are
​

Answer 1

Answer:

this answer is from work and energy chapter of class 11

Answer 2

Solution :-

So according to the given data we can write:

2|1i+xj+3k|=|4i+(4x−2)+2k|

We know that i is the unit vector, and the value of

i=1

So by breaking the mod and then by squaring the value we have:

2√{(1)²+(x)²+(3)²}=√{(4)²+(4x−2)²+(2)²}

WE can write (4x−2)²=(4x−2)(4x−2)

We can break down the terms and multiply the polynomial i.e.

4x(4x−2)−2(4x−2)=16x²−8x−8x+4

It gives us the value 16x²−16x+4.

Now back to the expression, we have:

2√(1+x²+9)=√(16+(4x−2)2+4)

We will square both the sides of the equation:

4(10+x²)=20+(4x−2)²

By putting the value of (4x−2)², we can wrote the above as:

40+4x²=20+16x²−16x+4

WE will transfer all the terms to the left hand side of the equation i.e.

40+4x²−20−16x²+16x−4=0

We will add the similar terms and we have:

−12x²−16x+16=0

Now we will divide the LHS by 4:

(−12x²−16x+16)/4=0

We can simplify the above as

4(−3x²−4x+4)/4=0

⇒−3x²−4x+4=0

Now we have a quadratic equation and we will use the middle term factorisation to solve this. We have an equation:

−3x²+4x+4=0

By taking the negative term out and without changing its meaning, it can also be written as

3x²−4x−4=0

We can write

3x²−4x−4=0⇒3x²−6x+2x−4=0

We will take the common factors out and it can be written as:

3x(x−2)+2(x−2)=0

We have now two factor here:

(x−2)(3x+2)

It gives us value

x−2=0⇒{{x=2}}

⇒3x+2=0

∴ {{x=−2/3}}