First slide

Straight lines in 3D

Question

The point in which the line _{$\frac{x + 1}{- 1} = \frac{y - 12}{5} = \frac{z - 7}{2}$} cuts the surface _{$11 x^{2} - 5 y^{2} - z^{2} = 0$} is

Easy

A

_{$(2, - 3, 1)$}
B

_{$(2, 3, - 1)$}
C

_{$(1, 2, 3)$}
D

_{$(1, 2, - 3)$}

Solution

Let _{$\frac{x + 1}{- 1} = \frac{y - 12}{5} = \frac{z - 7}{2}$} =r any point5 on the line is _{$(- r - 1, 5 r + 12, 2 r + 7)$} for every value of r. if this point lies on the surface
_{$11 x^{2} - 5 y^{2} - z^{2} = 0$} then _{$11 {(- r - 1)}^{2} - 5 {(5 r + 12)}^{2} + {(2 r + 7)}^{2} = 0$}

i.e. _{$11 r^{2} + 550 r + 660 = 0$} i.e. _{$r^{2} + 5 r + 6 = 0$}

i.e. _{$(r + 3) (r + 2) = 0$} i.e. _{$r = - 3, - 2$}

for these two values of r, the two points in which the given line cuts the surface are _{$(2, - 3, 1)$} and _{$(1, 2, 3)$}

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