Question

The equation of the plane through the line of intersection of planes ax + by + cz + d = 0,a'x + b'y + c'z+d' = 0 and parallel to the line y = 0,z = 0 is:

• A. (ab'-a'b)x+(bc'-b'c)y+(ad'- ad') = 0
• B. (ab'-a'b)x+(bc'-b'c) y+ (ad'-ad') z = 0
• C. (ab'-a'b)y+(ac'-a'c)z+(ad'-a'd) = 0
• D. (ab'-a'b)x+(ac'-a'c)z+(ad'-a'd) = 0

Answer 1

Answer: (C)

(ab'-a'b)y+(ac'-a'c)z+(ad'-a'd) = 0

Answer 2

The equation of a plane passing through the line of intersection of two planes $P_1 = ax + by + cz + d = 0$ and $P_2 = a'x + b'y + c'z + d' = 0$ is given by $P_1 + \lambda P_2 = 0$. This can be written as: $(ax + by + cz + d) + \lambda(a'x + b'y + c'z + d') = 0$ Rearranging terms, we get: $(a + \lambda a')x + (b + \lambda b')y + (c + \lambda c')z + (d + \lambda d') = 0$ The line $y=0, z=0$ is the x-axis, which has a direction vector $\vec{v} = (1, 0, 0)$. For the plane to be parallel to this line, its normal vector must be perpendicular to the direction vector of the line. The normal vector of the plane is $\vec{n} = (a + \lambda a', b + \lambda b', c + \lambda c')$. The condition $\vec{n} \cdot \vec{v} = 0$ is: $(a + \lambda a')(1) + (b + \lambda b')(0) + (c + \lambda c')(0) = 0$ $a + \lambda a' = 0$ This implies that the coefficient of $x$ in the equation of the plane must be zero. If $a' \neq 0$, then $\lambda = -a/a'$. Substituting this value of $\lambda$ into the plane equation: $(a + (-a/a')a')x + (b + (-a/a')b')y + (c + (-a/a')c')z + (d + (-a/a')d') = 0$ $(a - a)x + \left(b - \frac{ab'}{a'}\right)y + \left(c - \frac{ac'}{a'}\right)z + \left(d - \frac{ad'}{a'}\right) = 0$ $0x + \frac{a'b - ab'}{a'}y + \frac{a'c - ac'}{a'}z + \frac{a'd - ad'}{a'} = 0$ Multiplying the entire equation by $a'$ (assuming $a' \neq 0$): $(a'b - ab')y + (a'c - ac')z + (a'd - ad') = 0$ This equation can be rewritten by multiplying by $-1$: $-(a'b - ab')y - (a'c - ac')z - (a'd - ad') = 0$ $(ab' - a'b)y + (ac' - a'c)z + (ad' - a'd) = 0$