Question

Find the points where the planez= 5x−4y+ 3 intersects each of the coordinate axes. Findthe lengths of the sides and the angles of the triangle formed by these points.

Answer 1

You are given the plane $5x-4y-z+3=0.$

1.  To find x-intercept, you have to substitute y=z=0 into the plane's equation:

$5x-4\cdot 0-0+3=0,\\ \\x=-\dfrac{3}{5}.$

2. To find y-intercept, you have to substitute x=z=0 into the plane's equation:

$5\cdot 0-4y-0+3=0,\\ \\y=\dfrac{3}{4}.$

3. To find z-intercept, you have to substitute y=z=0 into the plane's equation:

$5\cdot 0-4\cdot 0-z+3=0,\\ \\z=3.$

You get three points of intersection with coordinate axises:

$A\left(-\dfrac{3}{5},0,0\right),\ B\left(0,\dfrac{3}{4},0\right),\ C(0,0,3).$

The lengths of sides are:

$OA=\dfrac{3}{5},\\ \\OB=\dfrac{3}{4},\\ \\OC=3,\\ \\AB=\sqrt{\left(\dfrac{3}{5}\right)^2+\left(\dfrac{3}{4}\right)^2}=3\sqrt{\dfrac{1}{25}+\dfrac{1}{16}}=\dfrac{3\sqrt{41} }{20},\\ \\BC=\sqrt{\left(\dfrac{3}{4}\right)^2+3^2}=3\sqrt{\dfrac{1}{16}+1}=\dfrac{3\sqrt{17} }{4},\\ \\AC=\sqrt{\left(\dfrac{3}{5}\right)^2+3^2}=3\sqrt{\dfrac{1}{25}+1}=\dfrac{3\sqrt{26} }{5}.$

To find angles consider vectors:

$\overrightarrow{AB}=\left(\dfrac{3}{5},\dfrac{3}{4},0\right),\\ \\\overrightarrow{AC}=\left(\dfrac{3}{5},0,3\right),\\ \\\overrightarrow{BC}=\left(0,-\dfrac{3}{4},3\right),\\ \\\overrightarrow{BA}=\left(-\dfrac{3}{5},-\dfrac{3}{4},0\right),\\ \\\overrightarrow{CA}=\left(-\dfrac{3}{5},0,-3\right),\\ \\\overrightarrow{CB}=\left(0,\dfrac{3}{4},-3\right).$

Then

$\cos \angle BAC=\dfrac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}|\cdot |\overrightarrow{AC}|}=\dfrac{\dfrac{9}{25}}{\dfrac{3\sqrt{26} }{5}\cdot \dfrac{3\sqrt{41} }{20}}=\dfrac{4}{\sqrt{26}\cdot \sqrt{41}}\Rightarrow \\ \\m\angle BAC=\arccos \left(\dfrac{4}{\sqrt{26}\cdot \sqrt{41}}\right)\approx 82.96^{\circ},$

$\cos \angle ABC=\dfrac{\overrightarrow{BA}\cdot \overrightarrow{BC}}{|\overrightarrow{BA}|\cdot |\overrightarrow{BC}|}=\dfrac{\dfrac{9}{16}}{\dfrac{3\sqrt{41} }{20}\cdot \dfrac{3\sqrt{17} }{4}}=\dfrac{5}{\sqrt{41}\cdot \sqrt{17}}\Rightarrow \\ \\m\angle ABC=\arccos \left(\dfrac{5}{\sqrt{41}\cdot \sqrt{17}}\right)\approx 79.08^{\circ},$

$\cos \angle BCA=\dfrac{\overrightarrow{CA}\cdot \overrightarrow{CB}}{|\overrightarrow{CA}|\cdot |\overrightarrow{CB}|}=\dfrac{9}{\dfrac{3\sqrt{26} }{5}\cdot \dfrac{3\sqrt{17} }{4}}=\dfrac{20}{\sqrt{26}\cdot \sqrt{17}}\Rightarrow \\ \\m\angle BCA=\arccos \left(\dfrac{20}{\sqrt{26}\cdot \sqrt{17}}\right)\approx 17.96^{\circ}$