Show that lines x = y, z = 0 and x + y = 0, z = 0 intersect each other

1.	Show that lines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.
Answer» Given lines are x = y, z = 0 and x + y = 0, z = 0. It is clear that (0, 0, 0) satisfies both the equations. ∴ the lines intersect at O whose position vector is \(\bar{0}\) Since z = 0 for both the lines, both the lines lie in XY- plane. Hence, we have to find equation of XY-plane. Z-axis is perpendicular to XY-plane. ∴ normal to XY plane is \(\hat{k}.\) \(0(\bar{0})\) lies on the plane. By using \(\bar{r} \) . \(\bar{n}\) = \(\bar{a}\) . \(\bar{n}\), the vector equation of the required plane is \(\bar{r} \) . \(\hat{k}\) = \(\bar{0}\) . \(\bar{k}\) i.e. \(\bar{r} \) . \(\hat{k}\) = 0. Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \) . \(\hat{k}\) = 0.

Show that lines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.

Answer»

Given lines are x = y, z = 0 and x + y = 0, z = 0.

It is clear that (0, 0, 0) satisfies both the equations.

∴ the lines intersect at O whose position vector is \(\bar{0}\)

Since z = 0 for both the lines, both the lines lie in XY- plane.

Hence, we have to find equation of XY-plane. Z-axis is perpendicular to XY-plane.

∴ normal to XY plane is \(\hat{k}.\)

\(0(\bar{0})\) lies on the plane.

By using \(\bar{r} \) . \(\bar{n}\) = \(\bar{a}\) . \(\bar{n}\), the vector equation of the required plane is \(\bar{r} \) . \(\hat{k}\) = \(\bar{0}\) . \(\bar{k}\)

i.e. \(\bar{r} \) . \(\hat{k}\) = 0.

Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \) . \(\hat{k}\) = 0.

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