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Q.

If a point R (4, y, z) lies on the line segment joining the points P (2, −3,4) and Q (8,0,10), then the distance of R from the origin is ____.

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Detailed Solution

Now consider the points P(2,−3,4) and Q(8,0,10) .
Now, we want to find the equation of line passing through the above points.
Now we have the equation of line passing through (x1,y1,z1) and (x2,y2,z2) is given by

\frac{x - x_{1}}{x_{1} - x_{2}} = \frac{y - y_{1}}{y_{1 -} y_{2}} = \frac{z - z_{1}}{z_{1} - z_{2}}

Hence the equation of line passing through P(2,−3,4)and Q(8,0,10) is

\frac{x - 2}{2 - 8} = \frac{y - (- 3)}{- 3 - 0} = \frac{z - 4}{4 - 10} \frac{x - 2}{- 6} = \frac{y + 3}{- 3} = \frac{z - 4}{- 6} = b

\frac{x - 2}{- 6} = b

x - 2 = - 6 b

x = - 6 b + 2

\frac{y + 3}{- 3} = b

y + 3 = - 3 b

y = - 3 b - 3

\frac{z - 4}{- 6} = b

z - 4 = - 6 b

z = - 6 b + 4

If a point R (4, y, z) lies on the line segment joining the points P (2, −3,4) and Q (8,0,10)
then a point R (4, y, z) lies on the line segment joining the points

(- 6 b + 2, - 3 b - 3, - 6 b + 4)

compare we get

- 6 b + 2 = 4

- 6 b = 4 - 2

b = \frac{2}{- 6} = \frac{- 1}{3}

(- 6 b + 2, - 3 b - 3, - 6 b + 4) = (- 6 (\frac{- 1}{3}) + 2, - 3 (\frac{- 1}{3}) - 3, - 6 (\frac{- 1}{3}) + 4)

(- 6 b + 2, - 3 b - 3, - 6 b + 4) = (4, - 2, 6)

Hence we get the point R is (4, - 2, 6)

Now we know coordinate of origin is (0, 0, 0)
Now distance between two points(

x_{1}, y_{1}, z_{1}) = (0, 0, 0,)

and(

x_{2}, y_{2}, z_{2}) = (4, - 2, 6)

is given by

\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} + {(z_{1} - z_{2})}^{2}} = \sqrt{{(0 - 4)}^{2} + {(0 + 2)}^{2} + {(0 - 6)}^{2}}

\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} + {(z_{1} - z_{2})}^{2}} = \sqrt{{(- 4)}^{2} + {(2)}^{2} + {(- 6)}^{2}}

\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} + {(z_{1} - z_{2})}^{2}} = \sqrt{16 + 4 + 36} = \sqrt{56} = \sqrt{14 * 4} = 2 \sqrt{14}

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