Question

Find an equation of the plane. The plane that passes through the point (3, 4, 5) and contains the line x = 5t, y = 3 + t, z = 4 − t

Answer 1

Answer: Hence the equation of plane is  x-4y+z+8=0

Explanation:

Since the points (3,4,5) and (0,3,4) are on the same plane

Now,

the vector a using two points (5,4,3) and (0,3,4) will be < -5,1,1>

and the vector b using points (5,4,3)and (3,4,5)  will be <-2,0,2>

Now, n=a×b

Normal vector (n) $= \begin{vmatrix}i &j &k\\-5&-1&1\\-2&0& 2 \end{vmatrix}$

n =-2i+8j-2k

Now the general equation of  plane:

$n.(x-x_0,y-y_0,z-z_0)=0$

$<-2,8,-2>\cdot <x-3,y-4,z-5>=0$

Equation of plane:

$-2(x-3)+8(y-4)-2(z-5)=0\\-2x+6+8y-32-2z+10=0$

$x-4y+z+8=0$

Hence the equation of plane is x-4y+z+8=0

Answer 2

Solution-

First of all we need co-ordinates of three points on the plane, to get the equation of that plane. We already have co-ordinate of one point on the plane, and we can get the co-ordinates of other two point from the given line equation. After getting all the 3 required co-ordinates, we can calculate two vectors on the plane and then taking their cross product will give us a normal to the desired plane.

Then we using the general equation for a plane with normal vector $\hat n= <a,b,c>$,

$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$

Putting t = 0, we get a point (0, 3, 4) which is also on the plane.

Putting t = 1, we get another point (5, 4, 3) which is also on the plane.

Let a be the vector from (0, 3, 4) to (3, 4, 5)

⇒ a = < 3-0, 4-3, 5-4 > = < 3, 1 , 1 >

Let b be the vector from (5, 4, 3) to (3, 4, 5)

⇒ b = < 3-5, 4-4, 5-3 > = < -2, 0, 2 >

Taking their cross product to get the normal,

$\left[\begin{array}{ccc}\hat i&\hat j&\hat k\\3&1&1\\-2&0&2\end{array}\right]$

= < 2, -8, 2>

∴ Then the equation for the plane passing through (3, 4, 5) and having normal vector < 2, -8, 2 > is,

$\Rightarrow 2(x-3)-8(y-4)+2(z-5)=0$

$\Rightarrow2x-8y+2z+16=0$