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=-(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=(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(-(3)/(5),0,0\right),\ B\left(0,(3)/(4),0\right),\ C(0,0,3).`

The lengths of sides are:

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

To find angles consider vectors:

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

Then

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

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

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