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

Vedclass · Accepted Answer

$51x + 15y - 50z + 173 = 0$

Vedclass · Answer

(C) The equation of any plane passing through the intersection of the planes $P_1: x + 2y + 3z - 4 = 0$ and $P_2: 2x + y - z + 5 = 0$ is given by $P_1 + \lambda P_2 = 0$. $(x + 2y + 3z - 4) + \lambda (2x + y - z + 5) = 0$ $(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (5\lambda - 4) = 0$ Since this plane is perpendicular to the plane $5x + 3y + 6z + 8 = 0$,the dot product of their normal vectors is zero. Normal vectors are $\vec{n_1} = (1 + 2\lambda, 2 + \lambda, 3 - \lambda)$ and $\vec{n_2} = (5, 3, 6)$. $5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0$ $5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$ $7\lambda + 29 = 0 \implies \lambda = -\frac{29}{7}$ Substituting $\lambda = -\frac{29}{7}$ into the equation: $(x + 2y + 3z - 4) - \frac{29}{7}(2x + y - z + 5) = 0$ $7x + 14y + 21z - 28 - 58x - 29y + 29z - 145 = 0$ $-51x - 15y + 50z - 173 = 0$ $51x + 15y - 50z + 173 = 0$

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$,and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

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