A point P lies on a line passing through Q(1, | vedclass

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(A) The equation of the line passing through $Q(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{

Vedclass · Accepted Answer

$\sqrt{42}$ units

Vedclass · Answer

(A) The equation of the line passing through $Q(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$. Since point $P$ lies on this line, its coordinates can be represented as $P(\lambda+1, 4\lambda-2, 5\lambda+3)$. Since $P$ also lies on the plane $2x + 3y - 4z + 22 = 0$, we substitute the coordinates of $P$ into the plane equation: $2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$. Expanding this, we get $2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$. Simplifying, we get $-6\lambda + 6 = 0$, which implies $\lambda = 1$. Substituting $\lambda = 1$ back into the coordinates of $P$, we get $P(1+1, 4(1)-2, 5(1)+3) = P(2, 2, 8)$. The distance $PQ$ is calculated as $\sqrt{(2-1)^2 + (2 - (-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$ units.

$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$, then the length of the segment $PQ$ is:

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