2x+3y-5z=0

3x-y+2z=0

I have found the direction vector using the vector product. The answer in the book says the lines goes through point (2,1,1). I think the 0's are a misprint. How do I find that the line goes though that point?

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- May 19th 2012, 07:45 AMStuck ManLine of intersection of two planes
2x+3y-5z=0

3x-y+2z=0

I have found the direction vector using the vector product. The answer in the book says the lines goes through point (2,1,1). I think the 0's are a misprint. How do I find that the line goes though that point? - May 19th 2012, 09:04 AMPlatoRe: Line of intersection of two planes
- May 19th 2012, 09:09 AMStuck ManRe: Line of intersection of two planes
Is it possible to figure out what the constants should be?

- May 19th 2012, 09:46 AMPlatoRe: Line of intersection of two planes
- May 19th 2012, 10:03 AMStuck ManRe: Line of intersection of two planes
The zeroes. If I put the point (2,1,1) into the plane equations I get -4 for the first constant and 7 for the other.

- May 19th 2012, 11:20 AMearbothRe: Line of intersection of two planes
The planes $\displaystyle p_1: 2x+3y-z=2$ and $\displaystyle p_2: 3x-y+2z = 7$ have the line of interception

$\displaystyle l:\left \{ \begin{array}{l}x=-t+\frac{23}{11} \\ y=19t - \frac8{11} \\ z= 11t\end{array}\right.$ With $\displaystyle t=\frac1{11}$ you'll get the point $\displaystyle (2,1,1)$ - May 19th 2012, 11:48 AMStuck ManRe: Line of intersection of two planes
Yes, thanks.