The length of the perpendicular from the poin | vedclass

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(D) Let the line be $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1} = \lambda$. Any point $M$ on the line is $(10\lambda-3, -7\lambda+2, \lambda)$. The vec

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greater than $3$ but less than $4$

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(D) Let the line be $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1} = \lambda$. Any point $M$ on the line is $(10\lambda-3, -7\lambda+2, \lambda)$. The vector $\vec{PM} = (10\lambda-3-2, -7\lambda+2-(-1), \lambda-4) = (10\lambda-5, -7\lambda+3, \lambda-4)$. Since $\vec{PM}$ is perpendicular to the line with direction vector $\vec{v} = (10, -7, 1)$,their dot product is zero: $10(10\lambda-5) - 7(-7\lambda+3) + 1(\lambda-4) = 0$ $100\lambda - 50 + 49\lambda - 21 + \lambda - 4 = 0$ $150\lambda - 75 = 0 \Rightarrow \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ in $M$,we get $M = (10(\frac{1}{2})-3, -7(\frac{1}{2})+2, \frac{1}{2}) = (2, -1.5, 0.5)$. The length $PM = \sqrt{(2-2)^2 + (-1.5 - (-1))^2 + (0.5-4)^2} = \sqrt{0^2 + (-0.5)^2 + (-3.5)^2} = \sqrt{0.25 + 12.25} = \sqrt{12.5}$. Since $\sqrt{9} Thus,the length is greater than $3$ but less than $4$.

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

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