Question

3. What is the cosine of the angle between the plane x+y+z= 5 and any of the

coordinate planes? What about the cosine of the angle between x+y+z= 5 and

x+ 2y+ 3z= 5?

Answer 1

Angle between planes equals angle between normal vectors of these planes.

Plane normal vector has coordinates $\vec{n}$ = (1, 1, 1).(1*x+1*y+1*z=5)

Plane 0YZ (x=0) normal vector has coordinates $\vec{i}$ =(1, 0, 0).

$cos(\vec{n},\vec{i})=\frac{\vec{n}*\vec{i}}{|\vec{n}|*|\vec{i}|}=\frac{1*1+1*0+1*0}{\sqrt{1^2+1^2+1^2}*\sqrt{1^2+0^2+0^2}}=\frac{1}{\sqrt{3}}$

Plane 0XZ (y=0) normal vector has coordinates $\vec{j}=(0, 1, 0).$

$cos(\vec{n},\vec{j})=\frac{\vec{n}*\vec{j}}{|\vec{n}|*|\vec{j}|}=\frac{1*0+1*1+1*0}{\sqrt{1^2+1^2+1^2}*\sqrt{0^2+1^2+0^2}}=\frac{1}{\sqrt{3}}$

Plane 0XY (z=0) normal vector has coordinates $\vec{k}=(0, 0, 1).$

$cos(\vec{n},\vec{k})=\frac{\vec{n}*\vec{k}}{|\vec{n}|*|\vec{k}|}=\frac{1*0+1*0+1*1}{\sqrt{1^2+1^2+1^2}*\sqrt{0^2+0^2+1^2}}=\frac{1}{\sqrt{3}}$

Plane x+2y+3z=5 normal vector has coordinates $\vec{m}=(1, 2, 3).$

$cos(\vec{n},\vec{m})=\frac{\vec{n}*\vec{m}}{|\vec{n}|*|\vec{m}|}=\frac{1*1+1*2+1*3}{\sqrt{1^2+1^2+1^2}*\sqrt{1^2+2^2+3^2}}=\frac{6}{\sqrt{42}}$