The plane $/4+/4+/7=1$ intersects the $-$ , $ -$ , and $$- axes in points $, , $. Find the area of the triangle $Δ$.

Question

The plane $/4+/4+/7=1$ intersects the $-$ , $ -$ , and $-$ axes in points $, , $. Find the area of the triangle $Δ$.

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.

Answer 1

Points of intersection are

$$P=(4,0,0)\quad Q=(0,4,0)\quad R=(0,0,7)\quad$$

then we can find the base and hight and then the area.

Otherwise we can use cross product

$$S=\frac12 \left|\vec{RP}\times \vec{RQ}\right|$$

Answer 2

The process you're going through doesn't find the area of the correct triangle. Finding the x-, y-, and z-intercepts is actually quite simple, as they are the solutions to the equation of the plane with two of the three variables equal to $0$.

$x/4+0/4+0/7=1$ implies the x-intercept is $(4,0,0)$. Similarly, the y- and z-intercepts are $(0,4,0)$ and $(0,0,7)$, respectively.

Now you could use find $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ and take half the magnitude of their cross product to find the area of the triangle.

Answer 3

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?

Answer 4

Let $O$ be an origin.

Thus, $OP=OQ=4,$ $OR=7,$ $PR=RQ=\sqrt{65},$ $PQ=\sqrt{32}$ and $$S_{\Delta PQR}=\frac{1}{4}\sqrt{2(65^2+65\cdot32+65\cdot32)-65^2-32^2-65^2}=\sqrt{456}.$$

Answer 5

Intercepts on the axes are $(4,4,7)$.

The sides are $ 4\sqrt2, \sqrt{65},\sqrt {65}...$ because they are perpendicular. Pythagoras thm can be used.

Use formula $\sqrt{s(s-a)(s-b)(s-c)}$