Question

5. Let u = [0,1,1), v = (-5, -4,6.7), and P = (4.–5.6). In the following, when rounding numbers, round to 4 decimal places. (i) Find the parametric form of the equation of the plane P. containing P and with direction vectors u and v. (ii) Find the parametric form of the equation of one of the two planes that are parallel to P, and distance 1 away from P1.

Answer 1

ANSWER :

i) Find the parametric form of the equation of the plane $P_{1}$ containing P and with direction vectors u and v.

The plane contains thevectors $u=[0,1,1],v=[-5,-4,6.7]$ and the point P = (4,-5,6)

Let , $x=(x,y,z)$ be an any point on
B1
. then, vector $x-P=[x-4,y+5,z-6]$ lies on this plane.

Now given any two vectors on a plane, a third vector on the plane can be written as a linear combination of the first two. So, $x-P=[x-4,y+5,z-6]$ can be written as a linear combination of the vectors u,v. So, there are scalarss such that

$x-P=su+tv$

$\Rightarrow x-(4,-5,6)=s(0,1,1)+t(5-,-4,6.7)$

$\Rightarrow x=(4,-5,6)+s(0,1,1)+t(5-,-4,6.7)$

Therefore, the parametric equation of the plane is

B1

$x=(4,-5,6)+s(0,1,1)+t(5-,-4,6.7)$

ii)

Find the parametric form of the equation of one of the two planes that are parallel to $P_{1}$ and distance 1 way form $P_{1}$ .

The plane parallel to the previous plane will have the same direction vectors, but won't pass through P = (4,-5,6). So, this equation will be of the form.

$x=(a,b,c)+s(0,1,1)+t(5-,-4,6.7)$

Now , u $\times$ v is perpendicular to the plane , and

B1

$u\times v$

$=\left [ 0,1,1 \right ]\times \left [ -5,-4,6.7 \right ]$

$=\left [ 10.7,-5,5 \right ]$

Now a unit vector in the direction of $u\times v$ is

$\frac{1}{\sqrt{10.7^{2}+(-5)^{2}-5^{2} }}\left [ 10.7,-5,5 \right ]=\frac{10}{\sqrt{16449}}\left [ 10.,-5,5 \right ]$

So, a vector, 1 unit away from the point P, in a direction perpendicular to the plane , is

B1

$\left ( 4,-5,6 \right )+\frac{10}{\sqrt{16449}}\left [ 10.7,-5,5 \right ]\approx (4.8343,-5.3898,6.3898)$

So, the plane 1 unit away from contains the point (4.8343,-5.3898,6.3898) .So, the equation of the plane is

B1

$X=(4.8343,-5.3898,6.3898)+s(0,1,1)+t(-5,-4,6,7)$