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(B) The equation of any plane passing through the line of intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y - z + 4 = 0$ is given

Vedclass · Accepted Answer

$y - 3z + 6 = 0$

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(B) The equation of any plane passing through the line of intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y - z + 4 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0$$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$ ... $(i)$Since the plane is parallel to the $x$-axis,its normal vector must be perpendicular to the $x$-axis (which has direction vector $\vec{i} = (1, 0, 0)$).Therefore,the coefficient of $x$ must be zero:$1 + 2\lambda = 0 \implies \lambda = -\frac{1}{2}$Substituting $\lambda = -\frac{1}{2}$ into equation $(i)$:$(1 + 2(-\frac{1}{2}))x + (1 + 3(-\frac{1}{2}))y + (1 - (-\frac{1}{2}))z + (4(-\frac{1}{2}) - 1) = 0$$0x + (1 - \frac{3}{2})y + (1 + \frac{1}{2})z + (-2 - 1) = 0$$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$Multiplying by $-2$,we get $y - 3z + 6 = 0$.

The equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

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