If the three planes x = 5,2x - 5ay + 3z - 2 = | vedclass

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(C) The three planes pass through a common line. This means the family of planes passing through the intersection of any two planes must contain the t

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$\left( \frac{1}{5}, -\frac{8}{15} \right)$

Vedclass · Answer

(C) The three planes pass through a common line. This means the family of planes passing through the intersection of any two planes must contain the third plane. Consider the planes $P_1: x - 5 = 0$,$P_2: 2x - 5ay + 3z - 2 = 0$,and $P_3: 3bx + y - 3z = 0$. The equation of any plane passing through the intersection of $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$. $(x - 5) + \lambda(2x - 5ay + 3z - 2) = 0$ $(1 + 2\lambda)x - (5a\lambda)y + (3\lambda)z - (5 + 2\lambda) = 0$. Since this plane is the same as $P_3$,we compare the coefficients with $3bx + y - 3z = 0$. For the coefficients to be proportional: $\frac{1 + 2\lambda}{3b} = \frac{-5a\lambda}{1} = \frac{3\lambda}{-3} = \frac{-(5 + 2\lambda)}{0}$. From $\frac{3\lambda}{-3} = -\lambda$,we have the ratio $k = -\lambda$. From $\frac{-(5 + 2\lambda)}{0}$,the denominator must be zero,so $0 = 0$ is not possible unless the constant term is also zero. However,looking at the ratio $\frac{3\lambda}{-3} = -\lambda$,we equate: $-5a\lambda = 1 \implies a = -\frac{1}{5\lambda}$. $3\lambda = -3 \implies \lambda = -1$. Substituting $\lambda = -1$ into the ratio: $a = -\frac{1}{5(-1)} = \frac{1}{5}$. Now for $b$: $\frac{1 + 2(-1)}{3b} = -(-1) = 1 \implies \frac{-1}{3b} = 1 \implies b = -\frac{1}{3}$. Wait,let us re-evaluate the constant term. The plane $P_3$ has no constant term,so the plane $P_1 + \lambda P_2$ must have a constant term of $0$. $-(5 + 2\lambda) = 0 \implies \lambda = -\frac{5}{2}$. Now substitute $\lambda = -\frac{5}{2}$ into the other ratios: $\frac{1 + 2(-5/2)}{3b} = \frac{-5a(-5/2)}{1} = \frac{3(-5/2)}{-3} = 1$. $\frac{1 - 5}{3b} = 1 \implies \frac{-4}{3b} = 1 \implies b = -\frac{4}{3}$. $\frac{25a}{2} = 1 \implies a = \frac{2}{25}$. Re-checking the question,if $x=5$ is a plane,it is $x-5=0$. The intersection of $x=5$ and $2x-5ay+3z-2=0$ gives $2(5)-5ay+3z-2=0 \implies -5ay+3z+8=0$. Comparing with $3bx+y-3z=0$ at $x=5$: $15b+y-3z=0$. Since they share a line,$y-3z = -15b$ and $-5ay+3z = -8$. Adding these: $(1-5a)y = -8-15b$. For this to hold for all points on the line,coefficients must match: $a = 1/5$ and $b = -8/15$. Thus,$(a, b) = (1/5, -8/15)$.

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is:

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