The vector $\hat{i}+x \hat{j}+3 \hat{k}$ is rotated through an angle of $\cos^{-1} \frac{11}{14}$ and doubled in magnitude, then it becomes $4 \hat{i}+(4x-2)\hat{j}+2 \hat{k}$.The value of $x$ cannot be which of the following?
$(A)$ $-\frac{2}{3}$
$(B)$ $\frac{2}{3}$
$(C)$ $-\frac{20}{17}$
$(A)$ $2$
Could someone briefly explain the concept of rotation of vector?
On
Do you really mean $\vec{i}+ (x+ 3)\vec{j}$, as you wrote, or do you mean $\vec{i}+ x\vec{j}+ 3\vec{k}$? And about what axis is the rotation?
If you mean $\vec{i}+ (x+ 3)\vec{j}$, which lies in the xy-plane, do you mean rotation about the z-axis?
In that case, the rotation is the same as $\begin{bmatrix}cos(cos^{-1}(11/14) & -sin(cos^{-1}(11/14) \\ sin(cos^{-1}(11/14) & cos(cos^{-1}(11/14)\end{bmatrix}\begin{bmatrix}1 \\ x+ 3\end{bmatrix}$ $= \begin{bmatrix}\frac{11}{14} & -\frac{5\sqrt{3}}{14} \\ \frac{5\sqrt{3}}{14} & \frac{11}{14}\end{bmatrix}\begin{bmatrix}1 \\ x+ 3\end{bmatrix}$
On
What you can do is work with the scalar product of the two vectors : $$\vec a.\vec b = |a||b| cos \theta$$ $$(\hat i + x \hat j + 3 \hat k ) . ( 4 \hat i + (4x - 2) \hat j + 2 \hat k) = \sqrt {1 + x^2 + 9}.\sqrt{16 + 16x^2 + 4 - 8x + 4} . \frac{11}{14} $$ Find your $x$'s from this equation.
Another equation can be : $$2|a| = |b|$$ $$\sqrt {1 + x^2 + 9} =\sqrt{16 + 16x^2 + 4 - 8x + 4}$$
Find your $x$'s from this equation too. Hopefully you will get the answer.
Cheers!
Rotating a vector is the act of changing its direction. Without much detail, one says that if you rotate a vector then you make it point in a different direction.
For example in the plane, the vector $(1,0) $ that 'points right' could be rotated by $90°$ to yield the vector $(0, 1) $ that 'points up'.
When rotating you only change directions, not sizes, so the vector has the same magnitude after rotating. The answer to your problem comes when you take into account the fact that we doubled its size after rotating it.