I need to find a direction vector for a line parallel to a plane $x+y+z = 0$ and that have $45$ degrees with the plane $x-y = 0$
So, i've assumed the vector $\vec V_r = (a,b,c)$ and since it is parallel to the first plane, the product:
$$(a,b,c)\cdot(1,1,1) = 0$$
(where $(1,1,1)$ is the vector normal to the first plane).
And also, using the formula for the angle between a line and a plane, where $n$ is the normal vector for the second plane:
$$\sin(\alpha) = \frac{|\vec V_r\cdot\vec n|}{\|\vec V_r\| \|\vec n\|}$$
so:
$$\sin\Bigl(\frac{\pi}{4}\Bigr) = \frac{\bigl|(a,b,c)\cdot(1,-1,0)\bigr|}{\sqrt{a^2+b^2+c^2}\sqrt{2}}$$
Then I end up with two equations:
$$a + b + c = 0$$
$$\frac{1}{\sqrt{2}} = \frac {|a-b|}{\sqrt{a^2+b^2+c^2}\sqrt{2}}$$
But i'm not able to solve for $(a,b,c)$. Could somebody help me? Thanks 🙂
Best Answer
After "cross multiplying" and canceling the factor of $\sqrt{2}$ on each side, we have $$ \sqrt{a^2 + b^2 + c^2} = |a - b|. $$ Now, square both sides and remove $a^2$ and $b^2$ terms that appear on both sides, yielding $$ c^2 = -2ab. $$
From the original equation, we know that $a + b + c = 0$, so $$ b = -a - c. $$
At this point, we can make the simplifying assumption that $a = 1$. Why? (Hover to reveal the answer after you've tried to answer this yourself.)
Substitute $b = -1 - c$ into $c^2 = -2b$, producing $c^2 = 2 + 2c$ or $(c - 1)^2 = 3$. Therefore, $$ c = 1 \pm \sqrt{3}. $$
Now, $$ b = -1 - (1 \pm \sqrt{3}) = -2 \mp \sqrt{3}. $$
Putting this all together, their are two possible direction vectors: $$ \vec V_r = \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \mp \sqrt{3} \\ 1 \pm \sqrt{3} \end{bmatrix}. $$