find the equation of the line of intersection between two planes $4x +2y -3z +5=0$ and $3x -6y +2z -3 = 0$

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Q. find the equation of the line of intersection between two planes

4x +2y -3z +5=0

and

3x -6y +2z -3 = 0

Find Direction Vector: To find the line of intersection between two planes, we need to find a direction vector that is perpendicular to the normal vectors of both planes. The normal vector of the first plane is $(4, 2, -3)$ and the normal vector of the second plane is $(3, -6, 2)$ . We can find the direction vector by taking the cross product of these two normal vectors.
Calculate Cross Product: Calculate the cross product of the normal vectors $(4, 2, -3)$ and $(3, -6, 2)$ . Using the determinant method for cross product: $\begin{vmatrix} i & j & k \ 4 & 2 & -3 \ 3 & -6 & 2 \end{vmatrix}$ The cross product is: $(2*(-3) - (-3)*(-6))i - (4*2 - (-3)*3)j + (4*(-6) - 2*3)k$ = $(-6 - 18)i - (8 - (-9))j + (-24 - 6)k$ = $-24i + 1j - 30k$
Find Point on Planes: The direction vector of the line of intersection is therefore $d = (-24, 1, -30)$ . Now we need to find a point that lies on both planes to determine the line completely. We can do this by setting two of the variables to zero and solving for the third in one of the plane equations.
Check Point on Second Plane: Let's set $y = 0$ and $z = 0$ and solve for $x$ using the first plane's equation $4x + 2y - 3z + 5 = 0$ . Substituting $y$ and $z$ we get: $4x + 2(0) - 3(0) + 5 = 0$ $4x + 5 = 0$ $4x = -5$ $x = -\frac{5}{4}$
Check Point on Second Plane: Let's set $y = 0$ and $z = 0$ and solve for $x$ using the first plane's equation $4x + 2y - 3z + 5 = 0$ . Substituting $y$ and $z$ we get: $4x + 2(0) - 3(0) + 5 = 0$ $4x + 5 = 0$ $4x = -5$ $x = -\frac{5}{4}$ We have found a point that lies on the first plane: $z = 0$ $0$ . We need to check if this point also lies on the second plane by substituting the values into the second plane's equation $z = 0$ $1$ . Substituting $x$ , $y$ , and $z$ we get: $z = 0$ $5$ $z = 0$ $6$ $z = 0$ $7$ $z = 0$ $8$