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

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(B) Let the direction ratios of the common line be $\ell, m, n$.Since the line lies on the plane $x = 5$,its direction vector must be perpendicular to

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the direction ratios of the common line be $\ell, m, n$.Since the line lies on the plane $x = 5$,its direction vector must be perpendicular to the normal $(1, 0, 0)$. Thus,$\ell = 0$.Since the line lies on the other two planes,its direction vector must be perpendicular to their normals $(2, -5a, 3)$ and $(3b, 1, -3)$.For the plane $2x - 5ay + 3z - 2 = 0$,we have $2\ell - 5am + 3n = 0$. Since $\ell = 0$,we get $-5am + 3n = 0$,or $3n = 5am$.For the plane $3bx + y - 3z = 0$,we have $3b\ell + m - 3n = 0$. Since $\ell = 0$,we get $m - 3n = 0$,or $m = 3n$.Substituting $m = 3n$ into $3n = 5am$,we get $3n = 5a(3n)$.Assuming $n 
eq 0$,we have $3 = 15a$,which gives $a = \frac{1}{5}$.Now,the common line must also satisfy the equations of the planes. Since $x = 5$,the other two equations become $-5ay + 3z = 2 - 2(5) = -8$ and $y - 3z = -3b(5) = -15b$.Substituting $a = \frac{1}{5}$,the first equation becomes $-y + 3z = -8$,or $y - 3z = 8$.Comparing $y - 3z = 8$ and $y - 3z = -15b$,we get $-15b = 8$,so $b = -\frac{8}{15}$.Thus,$(a, b) = \left( \frac{1}{5}, -\frac{8}{15} ight)$.

If the three planes $x = 5, 2x - 5ay + 3z - 2 = 0$ and $3bx + y - 3z = 0$ contain a common line,then $(a, b)$ is equal to

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