Question

find the vector eq of a plane pasing thru points (2,1,-1) and (-1,3,4) and perpendicular to the plane x-2y+4z=10.

Also show that the plane thus obtained contains the line r=-i+3j+4k+ L(3i-3j-5j)

(L --> lambda)

quik answer plzzz....

Answer 1

Answer:

The equation of the plane is:

        18x + 17y + 4z = 49

The given line does NOT lie in the plane though.  But it looks like it's probably just a mistake typing the question.  The line:

       r=-i+3j+4k+ L(3i-2j-5j)

does lie in the plane.  See below.

Step-by-step explanation:

Let the equation of the plane be

ax + by + cz = d.

Since (2,1,-1) and (-1,3,4) are in the plane:

2a + b - c = d

-a + 3b + 4c = d

Since the plane is perpendicular to the plane x-2y+4z=10, the normals (a,b,c) and (1,-2,4) to the two planes are orthogonal, so

a - 2b + 4c = 0.

We must solve these three simultaneous linear equations for a,b,c.  Using row operations on an augmented matrix, the process goes as follows:

$\displaystyle\left[\begin{array}{ccc|c}1&-2&4&0\\-1&3&4&d\\2&1&-1&d\end{array}\right]\rightarrow\left[\begin{array}{ccc|c}1&-2&4&0\\0&1&8&d\\0&5&-9&d\end{array}\right]\rightarrow\left[\begin{array}{ccc|c}1&0&20&2d\\0&1&8&d\\0&0&-49&-4d\end{array}\right]$

At this point, it looks like it will be convenient to take d = 49.  Doing this and continuing, we have...

$\displaystyle\rightarrow\left[\begin{array}{ccc|c}1&0&20&98\\0&1&8&49\\0&0&-49&-4\times49\end{array}\right]\rightarrow\left[\begin{array}{ccc|c}1&0&20&98\\0&1&8&49\\0&0&1&4\end{array}\right]\rightarrow\left[\begin{array}{ccc|c}1&0&0&18\\0&1&0&17\\0&0&1&4\end{array}\right]$

Hence for d = 49, we have a = 18, b = 17, c = 4.

Therefore the equation of the plane is:

18x + 17y + 4z = 49               ... (1)

To show that a line r = u + λv lies in the plane, it suffices to show that u is in the plane and that v is orthogonal to the normal to the plane.

Here we have u = ( -1, 3, 4 ), v = ( 3, -3, -5 ) and the normal to the plane is n = (18, 17, 4).

Putting x = -1, y = 3, z = 4 into the LHS of equation (1) gives

18(-1) + 17(3) + 4(4) = -18 + 51 + 16 = 49

so this point u = ( -1, 3, 4 ) does lie in the plane.      ... (*)

Taking the dot product between v and n gives

n.v = 18(3) + 17(-3) + 4(-5) = 54 - 51 - 20 = -17 ≠ 0.

As this is not zero, the line does NOT lie in the plane after all!

It's probably just a mistake in typing the question.  It probably should say

    " ... the line r=-i+3j+4k+ L(3i-2j-5j)"

Then v = ( 3, -2, -5 ) and this gives

n.v = 18(3) + 17(-2) + 4(-5) = 54 - 34 - 20 = 0

Now n and v are orthogonal.  Together with (*), this means the line lies in the plane.