For real numbers and 0, if the point of intersection of the straight lines x-1=y-12=z-13 a…

Question

For real numbers $\alpha$ and $\beta
eq0$, if the point of intersection of the straight lines $\dfrac{x-\alpha}{1}=\dfrac{y-1}{2}=\dfrac{z-1}{3}$ and $\dfrac{x-4}{\beta}=\dfrac{y-6}{3}=\dfrac{z-7}{3}$ lies on the plane $x+2y-z=8$, then $\alpha-\beta$ is equal to:

Accepted Answer

First line is ($$\phi$$ + $$\alpha$$, 2$$\phi$$ + 1, 3$$\phi$$ + 1)

and second line is (q$$\beta$$ + 4, 3q + 6, 3q + 7)

For intersection

$$\phi$$ + $$\alpha$$ = q$$\beta$$ + 4 ...... (i)

2$$\phi$$ + 1 = 3q + 6 .... (ii)

3$$\phi$$ + 1 = 3q + 7 ...... (iii)

for (ii) \& (iii) $$\phi$$ = 1, q = -1

So, from (i) $$\alpha$$ + $$\beta$$ = 3

Now, point of intersection is ($$\alpha$$ + 1, 3, 4)

It lies on the plane.

Hence, $$\alpha$$ = 5 \& $$\beta$$ = -2

Solutions · 1