A line with direction ratios 1, -1, 2 interse | vedclass

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(B) Let $P = (2k_1, 3k_1, 3k_1-1)$ and $Q = (-k_2-1, k_2+2, 4k_2)$.
The vector $\vec{PQ} = (-k_2-1-2k_1, k_2+2-3k_1, 4k_2-3k_1+1)$.
Since the direct

Vedclass · Accepted Answer

$1014$

Vedclass · Answer

(B) Let $P = (2k_1, 3k_1, 3k_1-1)$ and $Q = (-k_2-1, k_2+2, 4k_2)$.
The vector $\vec{PQ} = (-k_2-1-2k_1, k_2+2-3k_1, 4k_2-3k_1+1)$.
Since the direction ratios of the line $PQ$ are $(1, -1, 2)$, the components of $\vec{PQ}$ must be proportional to $(1, -1, 2)$.
Thus, $\frac{-k_2-1-2k_1}{1} = \frac{k_2+2-3k_1}{-1} = \frac{4k_2-3k_1+1}{2} = \lambda$.
From the first two parts: $-k_2-1-2k_1 = -k_2-2+3k_1 \implies 5k_1 = 1 \implies k_1 = 1/5$.
Substituting $k_1 = 1/5$ into the first and third parts: $-k_2-1-2/5 = \frac{4k_2-3/5+1}{2} \implies -k_2 - 7/5 = 2k_2 + 1/5 \implies 3k_2 = -8/5 \implies k_2 = -8/15$.
Then $\vec{PQ} = \lambda(1, -1, 2)$. Since $\lambda = -k_2-1-2k_1 = 8/15 - 1 - 2/5 = (8-15-6)/15 = -13/15$.
$PQ^2 = \alpha^2 = \lambda^2(1^2 + (-1)^2 + 2^2) = (-13/15)^2(6) = (169/225) 	imes 6 = 1014/225$.
Therefore, $225\alpha^2 = 1014$.

$A$ line with direction ratios $1, -1, 2$ intersects the lines $\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$ and $\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$ at the points $P$ and $Q$, respectively. If the length of the line segment $PQ$ is $\alpha$, then $225\alpha^2$ is equal to:

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