Question

Find the values of a,b so that the planes x+2y+z=b,3x-5y+3z=1 and 2x+7y+az=8 intersect in one point?

Answer 1

Guys i need help here .......

Answer 2

I guess the solution is not unique...

Answer 3

Tell me how to find..

Answer 4

.....

Answer 5

Okay, just my 2 cents worth of thoughts, don't know if it really works. Let the 3 planes be (P1), (P2) & (P3) resp., the eq. of the intersection (we call it (l1))of (P1) and (P2) is <x,y,z>+<1,0,-1>t. Now in order for (l1) and (P3) to intersect, <1,0,-1>.<2,7,a> must be nonzero

Answer 6

:(

Answer 7

Is it not good enough?

Answer 8

didnt get it? Im new to this lesson...

Answer 9

Please tell which part still puzzles you so I can elaborate a bit more on it.

Answer 10

\[\begin{align}x+2y+z&=b\\
3x-5y+3z&=1\\
2x+7y+az&=8\end{align}\]
\[\begin{bmatrix}1&2&1\\3&-5&3\\2&7&a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}b\\1\\8\end{bmatrix}\]
\[\left[\begin{array}{ccc|c}1&2&1&b\\3&-5&3&1\\2&7&a&8\end{array}\right]\]

Answer 11

Can u xplain the condition to 3 plains to intersect at one point?

Answer 12

The plains must be linearly independent

Answer 13

Thus the determinant must be equal zero. is tht wht I have to do here?

Answer 14

i think there are lots of ways to approach this problem

Answer 15

wht r they?

Answer 16

well i think i would try and reduce this
\[\left[\begin{array}{ccc|c}1&2&1&b\\3&-5&3&1\\2&7&a&8\end{array}\right]\]to this\[\left[\begin{array}{ccc|c}1&0&0&x_1\\0&1&0&y_1\\0&0&1&z_3\end{array}\right]\]