Find the point where the line and the plane intersects

Question

The line L in $R^3$ goes through the origin. A direction vector for the line creates the angle $60^o$ against the positive x-axis and and the angle $45^o$ against the positive y-axis. L intersects with the plane $x + \sqrt{2}y-3z=6 $ in the point P. Find P.

So, I'm completely lost here. Don't know what to with the "angle-information", and can't find any good information online, so really basic hints on this would be greatly appreciated.

Answer 1

Hint:

The orienting vector of the line is a unit vector $\vec u=(u_1,u_2,u_3)^T$ such that:

$ u_1^2+u_2^2+u_3^2=1$ ( because it is unitary)

$(u_1,u_2,u_3)\cdot(1,0,0)=\frac{1}{2}$ ( because it form an angle go $60°$ with the $x$ axis)

$(u_1,u_2,u_3)\cdot(0,1,0)=\frac{\sqrt{2}}{2}$ ( because it form an angle go 45° with the y axis)

can you do from this?

Answer 2

HINT

The direction unitary vectors $\vec v$ for the line can be found by

• $\vec v \cdot \vec i= v_x \cos 60°=\frac12 v_x$
• $\vec v \cdot \vec j= v_y\cos 45°=\frac{\sqrt 2}{2}v_y$

then find the intersection.

Answer 3

From the description of the angles the line makes with the $x$ and $y$ axes, the unit vector of the direction of the line $L$ is $$\left(\begin{matrix}\frac 12\\ \frac{1}{\sqrt{2}} \\ k\end{matrix}\right)$$

Since this is a unit vector, we can square and add these components and get a total of $1$, which gives two possible values for $k$, namely $\pm \frac 12$

So the equation of the line $L$ is $$\underline{r}=\lambda\left(\begin{matrix}\frac 12\\ \frac{1}{\sqrt{2}} \\ \pm\frac 12\end{matrix}\right)$$

Solving his simultaneously with the plane gives two possible values of $\lambda$ i.e. $3$ or $5$ and therefore two possible points of intersection