Question

Consider the following planes. 4x − 3y + z = 1, 3x + y − 4z = 4(a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.)(x(t), y(t), z(t)) = ?(b) Find the angle between the planes. (Round your answer to one decimal place.)

Answer 1

Answer with Step-by-step explanation:

We are given that two planes $4x-3y+z=1$ and $3x+y-4z=4$

a.We have to find the parametric equation of two given planes.

We will find a vector equation of line of intersection of two planes and one point on the line.

Let $\vec{a}=4\hat{i}-3\hat{j}+\hat{k}$

and $\vec{b}=3\hat{i}+\hat{j}-4\hat{k}$

We are finding vector equation of line of intersection by cross product

$\vec{a}\times \vec{b}=\begin{vmatrix}i&j&k\\4&-3&1\\3&1&-4\end{vmatrix}$

$\vec{a}\times \vec{b}=v=11\hat{i}+19\hat{j}+13\hat{k}$

Now, we are finding a point on the line of intersection

Substitute z=0

Then we get,4x-3y=1 (equation I)and 3x+y=4(Equation II)

Equation II multiply by 3 and then adding

13x=13

$x=\frac{13}{13}=1$

Substitute x=1 in equation I then we get

$4(1)-3y=1$

$4-1=3y$

$3y=3$

$y=\frac{3}{3}=1$

The point is$r_0= (1,1,0)$

Now, parametric equation of line of intersection of the planes

$r=r_0+t v$

Substitute the values then we get

$r=\hat{i}+\hat{j}+t(11\hat{i}+19\hat{j}+13\hat{k})$

$r(t)=(1+11t)\hat{i}+(1+19t)\hat{j}+13t\hat{k}$

Parametric equation

$x(t)=1+11t$

$y(t)=1+19t$

$z=13t$

b.Angle between two planes

$cos\theta=\frac{a\cdot b}{\mid a\mid\cdot\mid b\mid }$

Substitute the values then we get

$cos\theta=\frac{(4i-3j+k)\cdot(3i+j-4k)}{\sqrt{4^2+(-3)^2+(1)^2}\cdot \sqrt{3^2+1^1+(-4)^2}}$

$cos\theta=\frac{12-3-4}{\sqrt{26}\cdot\sqrt{26}}$

$cos\theta=\frac{5}{26}=0.1923$

$\theta=cso^{-1}(0.1923)=78.9^{\circ}$

$\theta=78.9^{\circ}$