The vector equation of the plane passing thro | vedclass

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(C) The equation of the family of planes passing through the intersection of $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y + 4z - 5 = 0$ is given by $P_1

Vedclass · Accepted Answer

$\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$

Vedclass · Answer

(C) The equation of the family of planes passing through the intersection of $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y + 4z - 5 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(x + y + z - 1) + \lambda(2x + 3y + 4z - 5) = 0$$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0$.This plane is perpendicular to the plane $x - y + z = 0$. The condition for perpendicularity of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.Therefore,$(1 + 2\lambda)(1) + (1 + 3\lambda)(-1) + (1 + 4\lambda)(1) = 0$.$1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$.$1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.Substituting $\lambda = -\frac{1}{3}$ into the equation of the plane:$(1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 + 4(-\frac{1}{3}))z - (1 + 5(-\frac{1}{3})) = 0$.$(1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0$.$\frac{1}{3}x + 0y - \frac{1}{3}z + \frac{2}{3} = 0$.Multiplying by $3$,we get $x - z + 2 = 0$.In vector form,this is $\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

The vector equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which is perpendicular to the plane $x - y + z = 0$ is

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