For what value of "k" will the following set of planes intersect in a line?

x - 2y - z = 0
x + 9y - 5z = 0
kx - y + z = 0

Work:

Elimination to find "z":
x - 2y - z = 0
x + 9y - 5z = 0
------------------
-11y + 4z = 0
z = 11y/4

Sub "z" into "x - 2y - z = 0" to find "x":
x - 2y -(11y/4) = 0
x - 19y/4 = 0
x = 19y/4

Sub "x" and "z" into "x + 9y - 5z = 0" to find "y":
(19y/4) + 9y - 5(11y/4) = 0
(19y + 36y - 55y) / 4 = 0
0y/4 = 0

So frustrated...I don't know what to do for this question. Please help me (step by step solution).

Thanks.

To find the value of "k" for which the set of planes intersect in a line, we need to find the values of "x," "y," and "z" that satisfy all three equations.

Let's start by solving for "z" using the first two equations:

Equation 1: x - 2y - z = 0
Equation 2: x + 9y - 5z = 0

To eliminate "x," we can subtract Equation 1 from Equation 2:

Equation 2 - Equation 1:

(x + 9y - 5z) - (x - 2y - z) = 0
x + 9y - 5z - x + 2y + z = 0
11y - 4z = 0
11y = 4z
z = (11y) / 4

Now substitute this value of "z" into Equation 1 to solve for "x":

x - 2y - z = 0
x - 2y - (11y/4) = 0
x - (8y + 11y) / 4 = 0
x = (8y + 11y) / 4
x = 19y / 4

Now substitute the values of "x" and "z" into Equation 2 to solve for "y":

(x + 9y - 5z) = 0
(19y / 4) + 9y - 5(11y/4) = 0
(19y + 36y - 55y) / 4 = 0
0y = 0

From this equation, we can see that "y" cancels out, meaning it can be any value. Therefore, there is no restriction on "y" and the value of "k" that will make the set of planes intersect in a line is any value.

In summary, for any value of "k," the set of planes will intersect in a line.

Sure! Let's go step by step to solve this question.

Step 1: Start with the given set of planes:
x - 2y - z = 0
x + 9y - 5z = 0
kx - y + z = 0

Step 2: To find the value of "k" for which these planes intersect in a line, we need to eliminate one variable from the equations. Let's eliminate "z" using the first two equations.

Step 3: Multiply the first equation by 5 and the second equation by 1, so that the coefficients of "z" will cancel out when we add them together:

5(x - 2y - z) = 0
1(x + 9y - 5z) = 0

This simplifies to:
5x - 10y - 5z = 0
x + 9y - 5z = 0

Step 4: Add the two new equations together:
(5x - 10y - 5z) + (x + 9y - 5z) = 0 + 0
6x - y - 10z = 0

Step 5: Now, we have the equation:
6x - y - 10z = 0

Step 6: We want these planes to intersect in a line, which means the coefficients of "x", "y", and "z" in the equation must be proportional. So, we can write the equation as:
6x = y + 10z

Step 7: Comparing the equation with the equation kx - y + z = 0, we can say that:
k = 6
Therefore, the value of "k" for which these planes intersect in a line is 6.

I hope this step-by-step explanation helps you understand how to solve this type of problem. Let me know if you have any further questions!

let l be th line x+1/2=y+3/3=-z and let P be the plane 3x-2y+4z=-1. find the piont of intersection po of l and p

Find the intersection of the first two lines:

subtract the first from the second
11y - 4z = 0
11y = 4z
y/z = 4/11

let y=4 and z=11 (it satisfies the above ratio)
back in the first

x - 8 - 11 = 0
x = 19

since clearly (0,0,0) lies on all three planes, the line joining (0,0,0) and (19,4,11) is the intersection line of the first two planes.
But all 3 planes are supposed to intersect in the same line, so obviously (19,4,11) must also lie on the third plane.
so sub it in....
k(19) - 8 - 11 = 0
k = 19/19 =1

so k=1