Straight lines in 3D

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The line $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z - 1}{- 1}$ intersects the curve $x^{2} - y^{2} = a^{2}; z = 0$ then absolute value of $a$ is equal to

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Solution

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

Putting these values in $x^{2} - y^{2} = a^{2},$ we get $a^{2} = 16 i . e ., a = \pm 4$

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