If the lines and intersect, then the absolute value of is

Question

If the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the absolute value of $(β - α)$ is _______

Smart Achievers · Accepted Answer

(1)

Given the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect

So, point on first line (1,2,3) and point on second line (4,1,0)

Vector joining both points is $- 3 \hat{i} + \hat{j} + 3 \hat{k}$

Now vector along first line is $2 \hat{i} + 3 \hat{j} + α \hat{k}$

Also vector along second line is $5 \hat{i} + 2 \hat{j} + β \hat{k}$

Now these three vectors must be coplanar $\Rightarrow$ $| \begin{matrix} 2 & 3 & α \\ 5 & 2 & β \\ - 3 & 1 & 3 \end{matrix} |$ $= 0$

$\Rightarrow 2 (6 - β) - 3 (15 + 3 β) + α (11) = 0$

$\Rightarrow α - β = 3$

Solution

Q.
If the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the absolute value of $(β - α)$ is _______

1 3

2 - 3

3 13

4 - 13

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