The line ell passes through the point P(2, 1, | vedclass

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(B) Let the direction ratios of line $\ell$ be $\langle a, b, c \rangle$. Since $\ell$ is parallel to the plane $x + y + 2z = 18$, its normal vector $

Vedclass · Accepted Answer

$\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z - 2}{-2}$

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(B) Let the direction ratios of line $\ell$ be $\langle a, b, c \rangle$. Since $\ell$ is parallel to the plane $x + y + 2z = 18$, its normal vector $\vec{n} = \langle 1, 1, 2 \rangle$ is perpendicular to $\ell$. Thus, $a + b + 2c = 0 \implies b = -a - 2c$.Any point on line $L_2$ is $Q(3k - 2, -k - 1, k + 2)$. Since $\ell$ passes through $P(2, 1, 1)$ and $Q$, the direction ratios are $\langle 3k - 4, -k - 2, k + 1 \rangle$.Equating ratios: $\frac{3k - 4}{a} = \frac{-k - 2}{b} = \frac{k + 1}{c} = \lambda$.Substitute $b = -a - 2c$ into the perpendicularity condition: $a + (-a - 2c) + 2c = 0$, which is always true. Using the direction ratios: $a(3k-4) + b(-k-2) + 2c(k+1) = 0$ is not correct; rather, the vector $\vec{PQ}$ must be perpendicular to the normal $\vec{n}$.So, $1(3k - 4) + 1(-k - 2) + 2(k + 1) = 0 \implies 3k - 4 - k - 2 + 2k + 2 = 0 \implies 4k = 4 \implies k = 1$.The point $Q$ is $(3(1) - 2, -1 - 1, 1 + 2) = (1, -2, 3)$.The direction vector of $\ell$ is $\vec{PQ} = \langle 1 - 2, -2 - 1, 3 - 1 \rangle = \langle -1, -3, 2 \rangle$, which is proportional to $\langle 1, 3, -2 \rangle$.The line equation is $\frac{x - 2}{1} = \frac{y - 1}{3} = \frac{z - 1}{-2}$, which matches option $B$.

The line $\ell$ passes through the point $P(2, 1, 1)$ and is parallel to the plane $x + y + 2z = 18$. If line $\ell$ intersects the line $L_2: \frac{x + 2}{3} = \frac{y + 1}{-1} = \frac{z - 2}{1}$, then the equation of the line $\ell$ is:

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