Let A and B be two points on the line frac{x} | vedclass

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(B) The equation of the line is $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1} = \lambda$. Any point on this line is of the form $A(\lambda, \lambda, -\lam

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

Vedclass · Answer

(B) The equation of the line is $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1} = \lambda$. Any point on this line is of the form $A(\lambda, \lambda, -\lambda)$. Given that the distance $PA = \sqrt{3}$,we have $PA^2 = 3$. $(\lambda - 1)^2 + (\lambda - 1)^2 + (-\lambda - 1)^2 = 3$. $(\lambda^2 - 2\lambda + 1) + (\lambda^2 - 2\lambda + 1) + (\lambda^2 + 2\lambda + 1) = 3$. $3\lambda^2 - 2\lambda + 3 = 3$. $3\lambda^2 - 2\lambda = 0$. $\lambda(3\lambda - 2) = 0$. Thus,$\lambda = 0$ or $\lambda = \frac{2}{3}$. The points are $A(0, 0, 0)$ and $B(\frac{2}{3}, \frac{2}{3}, -\frac{2}{3})$. The distance $AB = \sqrt{(\frac{2}{3} - 0)^2 + (\frac{2}{3} - 0)^2 + (-\frac{2}{3} - 0)^2}$. $AB = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

Let $A$ and $B$ be two points on the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1}$. If the distance of point $P(1, 1, 1)$ from the points $A$ and $B$ is $\sqrt{3}$,then the distance between $A$ and $B$ is:

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