The three planes $4 y + 6 z = 5, 2 x + 3 y + 5 z = 5$ and $6 x + 5 y + 9 z = 10$ -

Meet in a point.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Have a line in common.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Form a triangular prism.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C Have a line in common.
Given, $4 y + 6 z = 5, 2 x + 3 y + 5 z = 5$ and $6 x + 5 y + 9 z = 10$
First check for the coplanarity of the equation, it they are coplanar, then they do not intersect at a point, they intersect in a line.
So evaluating the scalar triple product, we have
$(4 j + 6 k) . (2 i + 3 j + 5 k) \times (6 i + 5 j + 9 k)$
$= (4 j + 6 k) . (2 i + 12 j - 8 k) = 48 - 48 = 0$
So these planes are co-planar and they have line in common.

Suggest Corrections