The vector a=- i+2 j+ k is rotated through a right angle, passing through the y-axis in it…

Question

The vector $\vec a=-\hat i+2\hat j+\hat k$ is rotated through a right angle, passing through the $y$-axis in its way, giving $\vec b$. Then the projection of $3\vec a+\sqrt2\vec b$ on $\vec c=5\hat i+4\hat j+3\hat k$ is

Accepted Answer

Since the rotation passes through the $y$-axis, $\vec b$ lies in the plane of $\vec a$ and $\hat j$, perpendicular to... using $\vec b=\lambda\,\vec a\times(\vec a\times\hat j)$ and matching $|\vec b|=|\vec a|=\sqrt6$ gives $\lambda=\pm\dfrac1{\sqrt2}$; taking the branch making an acute angle with the $y$-axis:
$$\vec b = \sqrt2(\hat i+\hat j-\hat k)$$

$$3\vec a+\sqrt2\vec b = 3(-1,2,1)+\sqrt2\cdot\sqrt2(1,1,-1) = (-3+2,\,6+2,\,3-2) = (-1,8,1)$$

$$\text{Projection on }\vec c = \frac{(-1)(5)+8(4)+1(3)}{|\vec c|} = \frac{-5+32+3}{\sqrt{50}} = \frac{30}{5\sqrt2} = 3\sqrt2$$

Solutions · 1