could you pls help me on these?
find the distance of:
(a) point P(1,3,1) to the surface of 3x+7y-5z=0
(b) point P(7,6,3) to the straight line of r=6j+k-t(i-j-k)
(c) L1:r=5k+t(3j+k) to L2:r=i+j-3k+t(i+3j+k)
thanks! =)
to a)
The equation 3x + 7y - 5z = 0 describes a plane with it's normal vector $\displaystyle \vec n = (3,7,-5)$. Transform the equation such that the normal vector is an unit vector. Plug in the coordinates of the staionary vector of P. The result is the distance of P from the plane:
$\displaystyle d = \dfrac{3 \cdot 1 + 7 \cdot 3 - 5 \cdot 1}{\sqrt{3^2+7^2+5^2}}$
to b)
Determine the equation of a plane which contains P and which is perpendicular to the line. Calculate the coordinates of the point of intersection between the line and the plane. Calculate the distance between these two points.
$\displaystyle p: -x+y+z-2=0$
point of intersection $\displaystyle Q\left(-\tfrac53 , \tfrac{13}3 , -\tfrac23 \right)$
$\displaystyle d = \sqrt{\frac{35}3}$
to c)
Determine the equation of a plane which contains line L1 and is parallel to line L2. Then take any point of L2 and calculate it's distance to the plane as you've done in a).
$\displaystyle p: \left((0,3,1) \times (1,3,1)\right) ((x,y,z) - (0,0,5))=0$ with $\displaystyle P(1,1,-3) \in L_2$