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(C) The equation of any plane passing through the intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ is given by the fam

Vedclass · Accepted Answer

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

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(C) The equation of any plane passing through the intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ is given by the family of planes equation:$(ax + by + cz + d) + \lambda(a'x + b'y + c'z + d') = 0$Rearranging the terms,we get:$x(a + \lambda a') + y(b + \lambda b') + z(c + \lambda c') + (d + \lambda d') = 0 \quad \dots(i)$Since the plane is parallel to the line $y = 0, z = 0$,which is the $x$-axis,its direction ratios are $(1, 0, 0)$.The normal to the plane $(i)$ is perpendicular to the line,so the dot product of the normal vector $(a + \lambda a', b + \lambda b', c + \lambda c')$ and the direction vector of the line $(1, 0, 0)$ must be zero:$1(a + \lambda a') + 0(b + \lambda b') + 0(c + \lambda c') = 0$$a + \lambda a' = 0 \implies \lambda = -\frac{a}{a'}$Substituting $\lambda = -\frac{a}{a'}$ into equation $(i)$:$(ax + by + cz + d) - \frac{a}{a'}(a'x + b'y + c'z + d') = 0$$a'ax + a'by + a'cz + a'd - aa'x - ab'y - ac'z - ad' = 0$$(a'b - ab')y + (a'c - ac')z + (a'd - ad') = 0$Multiplying by $-1$,we get:$(ab' - a'b)y + (ac' - a'c)z + (ad' - a'd) = 0$

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

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