The length of the perpendicular from the poin | vedclass

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(D) Let the line be $L: \frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1} = \lambda$. Any point $Q$ on the line is given by $Q(10\lambda - 3, -7\lambda +

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greater than $3$ but less than $4$

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(D) Let the line be $L: \frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1} = \lambda$. Any point $Q$ on the line is given by $Q(10\lambda - 3, -7\lambda + 2, \lambda)$. The vector $\vec{PQ} = (10\lambda - 3 - 2, -7\lambda + 2 - (-1), \lambda - 4) = (10\lambda - 5, -7\lambda + 3, \lambda - 4)$. Since $\vec{PQ}$ is perpendicular to the line direction $\vec{v} = (10, -7, 1)$,their dot product is $0$: $10(10\lambda - 5) - 7(-7\lambda + 3) + 1(\lambda - 4) = 0$. $100\lambda - 50 + 49\lambda - 21 + \lambda - 4 = 0$. $150\lambda - 75 = 0 \implies \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ into $\vec{PQ}$: $\vec{PQ} = (10(0.5) - 5, -7(0.5) + 3, 0.5 - 4) = (0, -0.5, -3.5)$. The length $PQ = \sqrt{0^2 + (-0.5)^2 + (-3.5)^2} = \sqrt{0.25 + 12.25} = \sqrt{12.5}$. Since $\sqrt{9} Thus,the length is greater than $3$ but less than $4$.

The length of the perpendicular from the point $P(2, -1, 4)$ to the straight line $\frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1}$ is:

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