First slide

Straight lines in 3D

Question

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1}$ intersects the curve $x y = c^{2}, z = 0$ if c=

Moderate

A

$\pm 1$
B

$\pm \sqrt{3}$
C

$\pm \sqrt{5}$
D

$\pm \sqrt{7}$

Solution

we have $z = 0$ for the point, where the line intersects the curve.
therefore $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{0 - 1}{- 1}$
$\Rightarrow \frac{x - 2}{3} = 1 a n d \frac{y + 1}{2} = 1$
$\Rightarrow x = 5 a n d y = 1$

Putting these values in $x y = c^{2}$ , we get $5 = c^{2} \Rightarrow c = \pm \sqrt{5}$

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