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

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$\vec{r} \cdot(\hat{j}-3 \hat{k})+6=0$

Vedclass · Answer

(A) The equations of the given planes are:$P_1: x+y+z-1=0$$P_2: 2x+3y-z+4=0$The equation of any plane passing through the line of intersection of these two planes is given by $P_1 + \lambda P_2 = 0$:$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$Since this plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}$ must be perpendicular to the direction of the $x$-axis,which is $\hat{i} = (1, 0, 0)$.Therefore,the dot product of the normal vector and the direction vector of the $x$-axis is zero:$(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$Substituting $\lambda = -\frac{1}{2}$ into the equation of the plane:$(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$$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 $-2$:$y - 3z + 6 = 0$In vector form,this is $\vec{r} \cdot (0\hat{i} + 1\hat{j} - 3\hat{k}) + 6 = 0$,which is $\vec{r} \cdot (\hat{j} - 3\hat{k}) + 6 = 0$.

The equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis is:

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