Question

The vector $i + xj + 3k$ is rotated through an angle $\theta $ and doubled in magnitude, then it becomes $4i + (4x - 2) + 2k$. The value of $x$ are
A. $\left( {\dfrac{{ - 2}}{3},2} \right)$
B. $\left( {\dfrac{1}{3},2} \right)$
C. $\left( {\dfrac{2}{3},0} \right)$
D. $\left( {2,7} \right)$

Answer 1

Hint: In the above question we have been given the vector i.e.
$i + xj + 3k$. When it is rotated by $\angle \theta $ and its magnitude is doubled, the second vector changes. We can write this as: $2\left| {1 + xj + 3k} \right| = \left| {4i + (4x - 2) + 2k} \right|$. We will solve this by first breaking the mod and then we square both the sides to get the required value.

Complete step by step answer:
So according to the given data we can write:
$2\left| {1 + xj + 3k} \right| = \left| {4i + (4x - 2) + 2k} \right|$
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\sqrt {{{(1)}^2} + {{(x)}^2} + {{(3)}^2}} = \sqrt {{{(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.
$4x(4x - 2) - 2(4x - 2) = 16{x^2} - 8x - 8x + 4$
It gives us the value $16{x^2} - 16x + 4$.

Now back to the expression, we have:
$2\sqrt {1 + {x^2} + 9} = \sqrt {16 + {{(4x - 2)}^2} + 4} $
We will square both the sides of the equation:
 $4(10 + {x^2}) = 20 + {(4x - 2)^2}$
By putting the value of ${(4x - 2)^2}$, we can wrote the above as:
$40 + 4{x^2} = 20 + 16{x^2} - 16x + 4$
WE will transfer all the terms to the left hand side of the equation i.e.
$40 + 4{x^2} - 20 - 16{x^2} + 16x - 4 = 0$
We will add the similar terms and we have:
$ - 12{x^2} - 16x + 16 = 0$
Now we will divide the LHS by $4$:
$\dfrac{{ - 12{x^2} - 16x + 16}}{4} = 0$
We can simplify the above as
$\dfrac{{4( - 3{x^2} - 4x + 4)}}{4} = 0 \Rightarrow - 3{x^2} - 4x + 4 = 0$

Now we have a quadratic equation and we will use the middle term factorisation to solve this. We have an equation:
$ - 3{x^2} + 4x + 4 = 0$
By taking the negative term out and without changing its meaning, it can also be written as
$3{x^2} - 4x - 4 = 0$
We can write
\[3{x^2} - 4x - 4 = 0 \Rightarrow 3{x^2} - 6x + 2x - 4 = 0\]
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
$x - 2 = 0 \\
\Rightarrow x = 2$
$\Rightarrow 3x + 2 = 0 \\
\therefore x = \dfrac{{ - 2}}{3}$

Hence the correct option is A.

Note:We should note that we have to break the middle term in such a way that
$p + q = 4x$ and
$\Rightarrow pq = - 12{x^2}$
So we have
$ - 6x + 2x = - 4x,( - 6x \times 2x) = - 12{x^2}$
This is called the middle term factorisation.
We should note that the vector in the direction of the x- axis is $i$. The vector in the direction of the y- axis is $j$. And the unit vector in the direction of the z- axis is $k$.