The plane $𝑥/4+𝑦/4+𝑧/7=1$ intersects the $𝑥-$ , $ 𝑦-$ , and $𝑧-$ axes in points $𝑃, 𝑄, 𝑅$. Find the area of the triangle $Δ𝑃𝑄𝑅$.
So here's my attempt.
First I find the normal vector:
$\left\langle\frac{1}{4}+\frac{1}{4}+\frac{1}{7}\right\rangle$
And then I find its magnitude:
$\frac{\sqrt{57}}{14\sqrt{2}}$
And then I take a half to it because 1/2 base times height for triangle area.
But this isn't right. What am I doing wrong? Thank you in advance.
Best Answer
The points $P,Q,R$ are easily seen to be $P=(4,0,0),Q=(0,4,0),R=(0,0,7)$. By the symmetry in the $x-$ and $y-$coordinates, this is an isosceles triangle with base $PQ$ and height $TR$, where $T$ is the mid-point of $PQ$. Can you take it from there?